Show that the following limit is infinity
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I want to show that the limit of $$frac{1}{n^alpha cdot ln(n)}$$ is infinity where $alpha < 0$. Is there an elementary way to show this? I know that $n^alpha$ gets small faster then $ln(n)$ gets large but still I can't find a formal proof. Thanks!
Edit: I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet.
limits
edited Dec 9 '18 at 16:48
Namaste
1
asked Dec 9 '18 at 15:32
OmerOmer
4039
$endgroup$
add a comment |
$begingroup$
I want to show that the limit of $$frac{1}{n^alpha cdot ln(n)}$$ is infinity where $alpha < 0$. Is there an elementary way to show this? I know that $n^alpha$ gets small faster then $ln(n)$ gets large but still I can't find a formal proof. Thanks!
Edit: I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet.
limits
edited Dec 9 '18 at 16:48
Namaste
1
asked Dec 9 '18 at 15:32
OmerOmer
4039
$endgroup$
$begingroup$
You are looking for a solution only by sequence? You should add that to your question in order to make it clear.
$endgroup$
– gimusi
Dec 9 '18 at 15:45
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@gimusi Okay. sorry. I edited the post.
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– Omer
Dec 9 '18 at 15:48
2
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What properties of the $log$ function do you know, yet?
$endgroup$
– Martin Rosenau
Dec 9 '18 at 16:13
add a comment |
$begingroup$
I want to show that the limit of $$frac{1}{n^alpha cdot ln(n)}$$ is infinity where $alpha < 0$. Is there an elementary way to show this? I know that $n^alpha$ gets small faster then $ln(n)$ gets large but still I can't find a formal proof. Thanks!
Edit: I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet.
limits
edited Dec 9 '18 at 16:48
Namaste
1
asked Dec 9 '18 at 15:32
OmerOmer
4039
$endgroup$
I want to show that the limit of $$frac{1}{n^alpha cdot ln(n)}$$ is infinity where $alpha < 0$. Is there an elementary way to show this? I know that $n^alpha$ gets small faster then $ln(n)$ gets large but still I can't find a formal proof. Thanks!
Edit: I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet.
limits
limits
edited Dec 9 '18 at 16:48
Namaste
1
asked Dec 9 '18 at 15:32
OmerOmer
4039
edited Dec 9 '18 at 16:48
Namaste
1
asked Dec 9 '18 at 15:32
OmerOmer
4039
edited Dec 9 '18 at 16:48
Namaste
1
edited Dec 9 '18 at 16:48
Namaste
1
edited Dec 9 '18 at 16:48
Namaste
1
1
asked Dec 9 '18 at 15:32
OmerOmer
4039
asked Dec 9 '18 at 15:32
OmerOmer
4039
asked Dec 9 '18 at 15:32
OmerOmer
4039
4039
$begingroup$
You are looking for a solution only by sequence? You should add that to your question in order to make it clear.
$endgroup$
– gimusi
Dec 9 '18 at 15:45
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@gimusi Okay. sorry. I edited the post.
$endgroup$
– Omer
Dec 9 '18 at 15:48
2
$begingroup$
What properties of the $log$ function do you know, yet?
$endgroup$
– Martin Rosenau
Dec 9 '18 at 16:13
add a comment |
$begingroup$
You are looking for a solution only by sequence? You should add that to your question in order to make it clear.
$endgroup$
– gimusi
Dec 9 '18 at 15:45
$begingroup$
@gimusi Okay. sorry. I edited the post.
$endgroup$
– Omer
Dec 9 '18 at 15:48
2
$begingroup$
What properties of the $log$ function do you know, yet?
$endgroup$
– Martin Rosenau
Dec 9 '18 at 16:13
$begingroup$
You are looking for a solution only by sequence? You should add that to your question in order to make it clear.
$endgroup$
– gimusi
Dec 9 '18 at 15:45
$begingroup$
You are looking for a solution only by sequence? You should add that to your question in order to make it clear.
$endgroup$
– gimusi
Dec 9 '18 at 15:45
$begingroup$
@gimusi Okay. sorry. I edited the post.
$endgroup$
– Omer
Dec 9 '18 at 15:48
$begingroup$
@gimusi Okay. sorry. I edited the post.
$endgroup$
– Omer
Dec 9 '18 at 15:48
2
2
$begingroup$
What properties of the $log$ function do you know, yet?
$endgroup$
– Martin Rosenau
Dec 9 '18 at 16:13
$begingroup$
What properties of the $log$ function do you know, yet?
$endgroup$
– Martin Rosenau
Dec 9 '18 at 16:13
add a comment |
4 Answers
4
active
oldest
votes
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A classic proof in high school that $;limdfrac{log x}x=0$ consists in observing that, since $tgesqrt t$ for $tge 1$, we have
$$log x=int_1^xfrac{mathrm dt}tle int_1^xfrac{mathrm dt}{sqrt t}=2bigl(sqrt x-1bigr)<2sqrt x,$$
hence $;dfrac{log x}x<dfrac{2sqrt x}x=dfrac 2{sqrt x}$, which tends to $0$ as $x$ tends to $infty$.
Now , setting $alpha =-a;(a>0)$, we deduce that
$$frac{1}{n^alpha cdot log(n)}=frac{n^a}{log n}=afrac{n^a}{log(n^a)}longrightarrow acdot +infty=+infty$$
by the above result.
answered Dec 9 '18 at 15:59
BernardBernard
123k741117
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"I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
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– gimusi
Dec 9 '18 at 16:01
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How does the O.P. know about the log, then?
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– Bernard
Dec 9 '18 at 16:02
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@gimusi: I don't see whay, in your 2nd displayed equation, the limit is $e^b$. Where has $a$ gone?
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– Bernard
Dec 9 '18 at 16:51
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It seems you miscalculated $;sqrt[n]{e^{n^a}}$: it is equal to $;e^{n^{a-1}}$.
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– Bernard
Dec 9 '18 at 17:03
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What about the new version in your opinion?
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– gimusi
Dec 9 '18 at 18:16
|
show 2 more comments
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Attempt:
$a>0$.
$dfrac{n^a}{log n}= dfrac{exp(log n^a)}{log n}=$
$dfrac{exp (alog n)}{log n}.$
$z:= log n.$
Now consider $z rightarrow infty$.
$f(z)=dfrac{exp (az)}{z} gt (a^2/2!)z.$
Hence $lim_{z rightarrow infty} f(z)=infty.$
Used:
$e^x =1+x +x^2/2!...> x^2/2!$, $x >0$.
answered Dec 9 '18 at 17:46
Peter SzilasPeter Szilas
11.6k2822
$endgroup$
$begingroup$
That seems nice even if the OP is looking "for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
$endgroup$
– gimusi
Dec 9 '18 at 18:16
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Gimusi.True.Delete?Your thoughts?
$endgroup$
– Peter Szilas
Dec 9 '18 at 18:30
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My opinion is that you can let it of course. There are also other answers with more advanced approach than what requested by the asker, and your one is different. Moreover some other reader could be interested to it.
$endgroup$
– gimusi
Dec 9 '18 at 18:34
add a comment |
$begingroup$
With
$$ln n<n-1$$
then
$$dfrac{1}{n^aln n}>dfrac{1}{n-1}n^{-a}toinfty$$
for $-a>1$.
Edit:
Might be helpful:
$$ln x=int_1^xfrac{mathrm dt}tle int_1^xfrac{t^bmathrm dt}{t}=int_1^x t^{b-1}mathrm dt=frac{x^b-1}{b}<frac{x^b}{b}$$
for real $b>0$. Now let $b=dfrac{a}{2}$, then
$$dfrac{1}{n^aln n}>dfrac{1}{n^frac{a}{2}n^frac{a}{2}}toinfty$$
if $-a>0$.
answered Dec 9 '18 at 15:47
NosratiNosrati
26.5k62354
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You don't have the case $--1le a<0$.
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– Bernard
Dec 9 '18 at 16:01
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the rest is proved with integrals and isn't elementary.
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– Nosrati
Dec 9 '18 at 16:02
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For me it's elementary inasmuch as it only uses elementary properties of integrals, as seen in high school. Or the O.P. should give us the definition of the log he/she uses.
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– Bernard
Dec 9 '18 at 16:06
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Sorry fro the (maybe stupid) question, but the first inequality is assumed as a given?
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– gimusi
Dec 9 '18 at 19:19
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As you know that one is proven with calculus, Isn't that?
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– Nosrati
Dec 9 '18 at 19:21
add a comment |
$begingroup$
Preliminary results
We have that $forall a>0$
$$frac{e^{n}} {n^a} to infty$$
indeed by ratio test
$$frac{e^{n+1}}{(n+1)^a}frac {n^a} {e^n}to e>1$$
then from here we have that
$$frac{log n}{n} to 0 iff frac n {log n}to infty$$
indeed for any $a >0$
$$frac n {log n}ge a iff nge log n^a iff e^nge n^a$$
Proof of the main result
By $b=-alpha>0$ we have
$$frac{1}{n^alpha cdot ln n }=frac{n^b}{ln n}=bfrac{n^b}{bln n}=bfrac{n^b}{ln n^b}$$
and since for any $b>0 implies n^bto infty$ we have that
$$bfrac{n^b}{ln n^b}to infty implies frac{1}{n^alpha cdot ln n } to infty$$
answered Dec 9 '18 at 15:39
gimusigimusi
93k84594
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A classic proof in high school that $;limdfrac{log x}x=0$ consists in observing that, since $tgesqrt t$ for $tge 1$, we have
$$log x=int_1^xfrac{mathrm dt}tle int_1^xfrac{mathrm dt}{sqrt t}=2bigl(sqrt x-1bigr)<2sqrt x,$$
hence $;dfrac{log x}x<dfrac{2sqrt x}x=dfrac 2{sqrt x}$, which tends to $0$ as $x$ tends to $infty$.
Now , setting $alpha =-a;(a>0)$, we deduce that
$$frac{1}{n^alpha cdot log(n)}=frac{n^a}{log n}=afrac{n^a}{log(n^a)}longrightarrow acdot +infty=+infty$$
by the above result.
answered Dec 9 '18 at 15:59
BernardBernard
123k741117
$endgroup$
$begingroup$
"I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
$endgroup$
– gimusi
Dec 9 '18 at 16:01
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How does the O.P. know about the log, then?
$endgroup$
– Bernard
Dec 9 '18 at 16:02
$begingroup$
@gimusi: I don't see whay, in your 2nd displayed equation, the limit is $e^b$. Where has $a$ gone?
$endgroup$
– Bernard
Dec 9 '18 at 16:51
$begingroup$
It seems you miscalculated $;sqrt[n]{e^{n^a}}$: it is equal to $;e^{n^{a-1}}$.
$endgroup$
– Bernard
Dec 9 '18 at 17:03
$begingroup$
What about the new version in your opinion?
$endgroup$
– gimusi
Dec 9 '18 at 18:16
|
show 2 more comments
$begingroup$
A classic proof in high school that $;limdfrac{log x}x=0$ consists in observing that, since $tgesqrt t$ for $tge 1$, we have
$$log x=int_1^xfrac{mathrm dt}tle int_1^xfrac{mathrm dt}{sqrt t}=2bigl(sqrt x-1bigr)<2sqrt x,$$
hence $;dfrac{log x}x<dfrac{2sqrt x}x=dfrac 2{sqrt x}$, which tends to $0$ as $x$ tends to $infty$.
Now , setting $alpha =-a;(a>0)$, we deduce that
$$frac{1}{n^alpha cdot log(n)}=frac{n^a}{log n}=afrac{n^a}{log(n^a)}longrightarrow acdot +infty=+infty$$
by the above result.
answered Dec 9 '18 at 15:59
BernardBernard
123k741117
$endgroup$
$begingroup$
"I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
$endgroup$
– gimusi
Dec 9 '18 at 16:01
$begingroup$
How does the O.P. know about the log, then?
$endgroup$
– Bernard
Dec 9 '18 at 16:02
$begingroup$
@gimusi: I don't see whay, in your 2nd displayed equation, the limit is $e^b$. Where has $a$ gone?
$endgroup$
– Bernard
Dec 9 '18 at 16:51
$begingroup$
It seems you miscalculated $;sqrt[n]{e^{n^a}}$: it is equal to $;e^{n^{a-1}}$.
$endgroup$
– Bernard
Dec 9 '18 at 17:03
$begingroup$
What about the new version in your opinion?
$endgroup$
– gimusi
Dec 9 '18 at 18:16
|
show 2 more comments
$begingroup$
A classic proof in high school that $;limdfrac{log x}x=0$ consists in observing that, since $tgesqrt t$ for $tge 1$, we have
$$log x=int_1^xfrac{mathrm dt}tle int_1^xfrac{mathrm dt}{sqrt t}=2bigl(sqrt x-1bigr)<2sqrt x,$$
hence $;dfrac{log x}x<dfrac{2sqrt x}x=dfrac 2{sqrt x}$, which tends to $0$ as $x$ tends to $infty$.
Now , setting $alpha =-a;(a>0)$, we deduce that
$$frac{1}{n^alpha cdot log(n)}=frac{n^a}{log n}=afrac{n^a}{log(n^a)}longrightarrow acdot +infty=+infty$$
by the above result.
answered Dec 9 '18 at 15:59
BernardBernard
123k741117
$endgroup$
A classic proof in high school that $;limdfrac{log x}x=0$ consists in observing that, since $tgesqrt t$ for $tge 1$, we have
$$log x=int_1^xfrac{mathrm dt}tle int_1^xfrac{mathrm dt}{sqrt t}=2bigl(sqrt x-1bigr)<2sqrt x,$$
hence $;dfrac{log x}x<dfrac{2sqrt x}x=dfrac 2{sqrt x}$, which tends to $0$ as $x$ tends to $infty$.
Now , setting $alpha =-a;(a>0)$, we deduce that
$$frac{1}{n^alpha cdot log(n)}=frac{n^a}{log n}=afrac{n^a}{log(n^a)}longrightarrow acdot +infty=+infty$$
by the above result.
answered Dec 9 '18 at 15:59
BernardBernard
123k741117
edited Dec 9 '18 at 19:15
answered Dec 9 '18 at 15:59
BernardBernard
123k741117
answered Dec 9 '18 at 15:59
BernardBernard
123k741117
answered Dec 9 '18 at 15:59
BernardBernard
123k741117
123k741117
$begingroup$
"I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
$endgroup$
– gimusi
Dec 9 '18 at 16:01
$begingroup$
How does the O.P. know about the log, then?
$endgroup$
– Bernard
Dec 9 '18 at 16:02
$begingroup$
@gimusi: I don't see whay, in your 2nd displayed equation, the limit is $e^b$. Where has $a$ gone?
$endgroup$
– Bernard
Dec 9 '18 at 16:51
$begingroup$
It seems you miscalculated $;sqrt[n]{e^{n^a}}$: it is equal to $;e^{n^{a-1}}$.
$endgroup$
– Bernard
Dec 9 '18 at 17:03
$begingroup$
What about the new version in your opinion?
$endgroup$
– gimusi
Dec 9 '18 at 18:16
|
show 2 more comments
$begingroup$
"I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
$endgroup$
– gimusi
Dec 9 '18 at 16:01
$begingroup$
How does the O.P. know about the log, then?
$endgroup$
– Bernard
Dec 9 '18 at 16:02
$begingroup$
@gimusi: I don't see whay, in your 2nd displayed equation, the limit is $e^b$. Where has $a$ gone?
$endgroup$
– Bernard
Dec 9 '18 at 16:51
$begingroup$
It seems you miscalculated $;sqrt[n]{e^{n^a}}$: it is equal to $;e^{n^{a-1}}$.
$endgroup$
– Bernard
Dec 9 '18 at 17:03
$begingroup$
What about the new version in your opinion?
$endgroup$
– gimusi
Dec 9 '18 at 18:16
$begingroup$
"I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
$endgroup$
– gimusi
Dec 9 '18 at 16:01
$begingroup$
"I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
$endgroup$
– gimusi
Dec 9 '18 at 16:01
$begingroup$
How does the O.P. know about the log, then?
$endgroup$
– Bernard
Dec 9 '18 at 16:02
$begingroup$
How does the O.P. know about the log, then?
$endgroup$
– Bernard
Dec 9 '18 at 16:02
$begingroup$
@gimusi: I don't see whay, in your 2nd displayed equation, the limit is $e^b$. Where has $a$ gone?
$endgroup$
– Bernard
Dec 9 '18 at 16:51
$begingroup$
@gimusi: I don't see whay, in your 2nd displayed equation, the limit is $e^b$. Where has $a$ gone?
$endgroup$
– Bernard
Dec 9 '18 at 16:51
$begingroup$
It seems you miscalculated $;sqrt[n]{e^{n^a}}$: it is equal to $;e^{n^{a-1}}$.
$endgroup$
– Bernard
Dec 9 '18 at 17:03
$begingroup$
It seems you miscalculated $;sqrt[n]{e^{n^a}}$: it is equal to $;e^{n^{a-1}}$.
$endgroup$
– Bernard
Dec 9 '18 at 17:03
$begingroup$
What about the new version in your opinion?
$endgroup$
– gimusi
Dec 9 '18 at 18:16
$begingroup$
What about the new version in your opinion?
$endgroup$
– gimusi
Dec 9 '18 at 18:16
|
show 2 more comments
$begingroup$
Attempt:
$a>0$.
$dfrac{n^a}{log n}= dfrac{exp(log n^a)}{log n}=$
$dfrac{exp (alog n)}{log n}.$
$z:= log n.$
Now consider $z rightarrow infty$.
$f(z)=dfrac{exp (az)}{z} gt (a^2/2!)z.$
Hence $lim_{z rightarrow infty} f(z)=infty.$
Used:
$e^x =1+x +x^2/2!...> x^2/2!$, $x >0$.
answered Dec 9 '18 at 17:46
Peter SzilasPeter Szilas
11.6k2822
$endgroup$
$begingroup$
That seems nice even if the OP is looking "for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
$endgroup$
– gimusi
Dec 9 '18 at 18:16
$begingroup$
Gimusi.True.Delete?Your thoughts?
$endgroup$
– Peter Szilas
Dec 9 '18 at 18:30
$begingroup$
My opinion is that you can let it of course. There are also other answers with more advanced approach than what requested by the asker, and your one is different. Moreover some other reader could be interested to it.
$endgroup$
– gimusi
Dec 9 '18 at 18:34
add a comment |
$begingroup$
Attempt:
$a>0$.
$dfrac{n^a}{log n}= dfrac{exp(log n^a)}{log n}=$
$dfrac{exp (alog n)}{log n}.$
$z:= log n.$
Now consider $z rightarrow infty$.
$f(z)=dfrac{exp (az)}{z} gt (a^2/2!)z.$
Hence $lim_{z rightarrow infty} f(z)=infty.$
Used:
$e^x =1+x +x^2/2!...> x^2/2!$, $x >0$.
answered Dec 9 '18 at 17:46
Peter SzilasPeter Szilas
11.6k2822
$endgroup$
$begingroup$
That seems nice even if the OP is looking "for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
$endgroup$
– gimusi
Dec 9 '18 at 18:16
$begingroup$
Gimusi.True.Delete?Your thoughts?
$endgroup$
– Peter Szilas
Dec 9 '18 at 18:30
$begingroup$
My opinion is that you can let it of course. There are also other answers with more advanced approach than what requested by the asker, and your one is different. Moreover some other reader could be interested to it.
$endgroup$
– gimusi
Dec 9 '18 at 18:34
add a comment |
$begingroup$
Attempt:
$a>0$.
$dfrac{n^a}{log n}= dfrac{exp(log n^a)}{log n}=$
$dfrac{exp (alog n)}{log n}.$
$z:= log n.$
Now consider $z rightarrow infty$.
$f(z)=dfrac{exp (az)}{z} gt (a^2/2!)z.$
Hence $lim_{z rightarrow infty} f(z)=infty.$
Used:
$e^x =1+x +x^2/2!...> x^2/2!$, $x >0$.
answered Dec 9 '18 at 17:46
Peter SzilasPeter Szilas
11.6k2822
$endgroup$
Attempt:
$a>0$.
$dfrac{n^a}{log n}= dfrac{exp(log n^a)}{log n}=$
$dfrac{exp (alog n)}{log n}.$
$z:= log n.$
Now consider $z rightarrow infty$.
$f(z)=dfrac{exp (az)}{z} gt (a^2/2!)z.$
Hence $lim_{z rightarrow infty} f(z)=infty.$
Used:
$e^x =1+x +x^2/2!...> x^2/2!$, $x >0$.
answered Dec 9 '18 at 17:46
Peter SzilasPeter Szilas
11.6k2822
answered Dec 9 '18 at 17:46
Peter SzilasPeter Szilas
11.6k2822
answered Dec 9 '18 at 17:46
Peter SzilasPeter Szilas
11.6k2822
answered Dec 9 '18 at 17:46
Peter SzilasPeter Szilas
11.6k2822
11.6k2822
$begingroup$
That seems nice even if the OP is looking "for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
$endgroup$
– gimusi
Dec 9 '18 at 18:16
$begingroup$
Gimusi.True.Delete?Your thoughts?
$endgroup$
– Peter Szilas
Dec 9 '18 at 18:30
$begingroup$
My opinion is that you can let it of course. There are also other answers with more advanced approach than what requested by the asker, and your one is different. Moreover some other reader could be interested to it.
$endgroup$
– gimusi
Dec 9 '18 at 18:34
add a comment |
$begingroup$
That seems nice even if the OP is looking "for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
$endgroup$
– gimusi
Dec 9 '18 at 18:16
$begingroup$
Gimusi.True.Delete?Your thoughts?
$endgroup$
– Peter Szilas
Dec 9 '18 at 18:30
$begingroup$
My opinion is that you can let it of course. There are also other answers with more advanced approach than what requested by the asker, and your one is different. Moreover some other reader could be interested to it.
$endgroup$
– gimusi
Dec 9 '18 at 18:34
$begingroup$
That seems nice even if the OP is looking "for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
$endgroup$
– gimusi
Dec 9 '18 at 18:16
$begingroup$
That seems nice even if the OP is looking "for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
$endgroup$
– gimusi
Dec 9 '18 at 18:16
$begingroup$
Gimusi.True.Delete?Your thoughts?
$endgroup$
– Peter Szilas
Dec 9 '18 at 18:30
$begingroup$
Gimusi.True.Delete?Your thoughts?
$endgroup$
– Peter Szilas
Dec 9 '18 at 18:30
$begingroup$
My opinion is that you can let it of course. There are also other answers with more advanced approach than what requested by the asker, and your one is different. Moreover some other reader could be interested to it.
$endgroup$
– gimusi
Dec 9 '18 at 18:34
$begingroup$
My opinion is that you can let it of course. There are also other answers with more advanced approach than what requested by the asker, and your one is different. Moreover some other reader could be interested to it.
$endgroup$
– gimusi
Dec 9 '18 at 18:34
add a comment |
$begingroup$
With
$$ln n<n-1$$
then
$$dfrac{1}{n^aln n}>dfrac{1}{n-1}n^{-a}toinfty$$
for $-a>1$.
Edit:
Might be helpful:
$$ln x=int_1^xfrac{mathrm dt}tle int_1^xfrac{t^bmathrm dt}{t}=int_1^x t^{b-1}mathrm dt=frac{x^b-1}{b}<frac{x^b}{b}$$
for real $b>0$. Now let $b=dfrac{a}{2}$, then
$$dfrac{1}{n^aln n}>dfrac{1}{n^frac{a}{2}n^frac{a}{2}}toinfty$$
if $-a>0$.
answered Dec 9 '18 at 15:47
NosratiNosrati
26.5k62354
$endgroup$
$begingroup$
You don't have the case $--1le a<0$.
$endgroup$
– Bernard
Dec 9 '18 at 16:01
$begingroup$
the rest is proved with integrals and isn't elementary.
$endgroup$
– Nosrati
Dec 9 '18 at 16:02
$begingroup$
For me it's elementary inasmuch as it only uses elementary properties of integrals, as seen in high school. Or the O.P. should give us the definition of the log he/she uses.
$endgroup$
– Bernard
Dec 9 '18 at 16:06
$begingroup$
Sorry fro the (maybe stupid) question, but the first inequality is assumed as a given?
$endgroup$
– gimusi
Dec 9 '18 at 19:19
$begingroup$
As you know that one is proven with calculus, Isn't that?
$endgroup$
– Nosrati
Dec 9 '18 at 19:21
add a comment |
$begingroup$
With
$$ln n<n-1$$
then
$$dfrac{1}{n^aln n}>dfrac{1}{n-1}n^{-a}toinfty$$
for $-a>1$.
Edit:
Might be helpful:
$$ln x=int_1^xfrac{mathrm dt}tle int_1^xfrac{t^bmathrm dt}{t}=int_1^x t^{b-1}mathrm dt=frac{x^b-1}{b}<frac{x^b}{b}$$
for real $b>0$. Now let $b=dfrac{a}{2}$, then
$$dfrac{1}{n^aln n}>dfrac{1}{n^frac{a}{2}n^frac{a}{2}}toinfty$$
if $-a>0$.
answered Dec 9 '18 at 15:47
NosratiNosrati
26.5k62354
$endgroup$
$begingroup$
You don't have the case $--1le a<0$.
$endgroup$
– Bernard
Dec 9 '18 at 16:01
$begingroup$
the rest is proved with integrals and isn't elementary.
$endgroup$
– Nosrati
Dec 9 '18 at 16:02
$begingroup$
For me it's elementary inasmuch as it only uses elementary properties of integrals, as seen in high school. Or the O.P. should give us the definition of the log he/she uses.
$endgroup$
– Bernard
Dec 9 '18 at 16:06
$begingroup$
Sorry fro the (maybe stupid) question, but the first inequality is assumed as a given?
$endgroup$
– gimusi
Dec 9 '18 at 19:19
$begingroup$
As you know that one is proven with calculus, Isn't that?
$endgroup$
– Nosrati
Dec 9 '18 at 19:21
add a comment |
$begingroup$
With
$$ln n<n-1$$
then
$$dfrac{1}{n^aln n}>dfrac{1}{n-1}n^{-a}toinfty$$
for $-a>1$.
Edit:
Might be helpful:
$$ln x=int_1^xfrac{mathrm dt}tle int_1^xfrac{t^bmathrm dt}{t}=int_1^x t^{b-1}mathrm dt=frac{x^b-1}{b}<frac{x^b}{b}$$
for real $b>0$. Now let $b=dfrac{a}{2}$, then
$$dfrac{1}{n^aln n}>dfrac{1}{n^frac{a}{2}n^frac{a}{2}}toinfty$$
if $-a>0$.
answered Dec 9 '18 at 15:47
NosratiNosrati
26.5k62354
$endgroup$
With
$$ln n<n-1$$
then
$$dfrac{1}{n^aln n}>dfrac{1}{n-1}n^{-a}toinfty$$
for $-a>1$.
Edit:
Might be helpful:
$$ln x=int_1^xfrac{mathrm dt}tle int_1^xfrac{t^bmathrm dt}{t}=int_1^x t^{b-1}mathrm dt=frac{x^b-1}{b}<frac{x^b}{b}$$
for real $b>0$. Now let $b=dfrac{a}{2}$, then
$$dfrac{1}{n^aln n}>dfrac{1}{n^frac{a}{2}n^frac{a}{2}}toinfty$$
if $-a>0$.
answered Dec 9 '18 at 15:47
NosratiNosrati
26.5k62354
edited Dec 9 '18 at 16:33
answered Dec 9 '18 at 15:47
NosratiNosrati
26.5k62354
answered Dec 9 '18 at 15:47
NosratiNosrati
26.5k62354
answered Dec 9 '18 at 15:47
NosratiNosrati
26.5k62354
26.5k62354
$begingroup$
You don't have the case $--1le a<0$.
$endgroup$
– Bernard
Dec 9 '18 at 16:01
$begingroup$
the rest is proved with integrals and isn't elementary.
$endgroup$
– Nosrati
Dec 9 '18 at 16:02
$begingroup$
For me it's elementary inasmuch as it only uses elementary properties of integrals, as seen in high school. Or the O.P. should give us the definition of the log he/she uses.
$endgroup$
– Bernard
Dec 9 '18 at 16:06
$begingroup$
Sorry fro the (maybe stupid) question, but the first inequality is assumed as a given?
$endgroup$
– gimusi
Dec 9 '18 at 19:19
$begingroup$
As you know that one is proven with calculus, Isn't that?
$endgroup$
– Nosrati
Dec 9 '18 at 19:21
add a comment |
$begingroup$
You don't have the case $--1le a<0$.
$endgroup$
– Bernard
Dec 9 '18 at 16:01
$begingroup$
the rest is proved with integrals and isn't elementary.
$endgroup$
– Nosrati
Dec 9 '18 at 16:02
$begingroup$
For me it's elementary inasmuch as it only uses elementary properties of integrals, as seen in high school. Or the O.P. should give us the definition of the log he/she uses.
$endgroup$
– Bernard
Dec 9 '18 at 16:06
$begingroup$
Sorry fro the (maybe stupid) question, but the first inequality is assumed as a given?
$endgroup$
– gimusi
Dec 9 '18 at 19:19
$begingroup$
As you know that one is proven with calculus, Isn't that?
$endgroup$
– Nosrati
Dec 9 '18 at 19:21
$begingroup$
You don't have the case $--1le a<0$.
$endgroup$
– Bernard
Dec 9 '18 at 16:01
$begingroup$
You don't have the case $--1le a<0$.
$endgroup$
– Bernard
Dec 9 '18 at 16:01
$begingroup$
the rest is proved with integrals and isn't elementary.
$endgroup$
– Nosrati
Dec 9 '18 at 16:02
$begingroup$
the rest is proved with integrals and isn't elementary.
$endgroup$
– Nosrati
Dec 9 '18 at 16:02
$begingroup$
For me it's elementary inasmuch as it only uses elementary properties of integrals, as seen in high school. Or the O.P. should give us the definition of the log he/she uses.
$endgroup$
– Bernard
Dec 9 '18 at 16:06
$begingroup$
For me it's elementary inasmuch as it only uses elementary properties of integrals, as seen in high school. Or the O.P. should give us the definition of the log he/she uses.
$endgroup$
– Bernard
Dec 9 '18 at 16:06
$begingroup$
Sorry fro the (maybe stupid) question, but the first inequality is assumed as a given?
$endgroup$
– gimusi
Dec 9 '18 at 19:19
$begingroup$
Sorry fro the (maybe stupid) question, but the first inequality is assumed as a given?
$endgroup$
– gimusi
Dec 9 '18 at 19:19
$begingroup$
As you know that one is proven with calculus, Isn't that?
$endgroup$
– Nosrati
Dec 9 '18 at 19:21
$begingroup$
As you know that one is proven with calculus, Isn't that?
$endgroup$
– Nosrati
Dec 9 '18 at 19:21
add a comment |
$begingroup$
Preliminary results
We have that $forall a>0$
$$frac{e^{n}} {n^a} to infty$$
indeed by ratio test
$$frac{e^{n+1}}{(n+1)^a}frac {n^a} {e^n}to e>1$$
then from here we have that
$$frac{log n}{n} to 0 iff frac n {log n}to infty$$
indeed for any $a >0$
$$frac n {log n}ge a iff nge log n^a iff e^nge n^a$$
Proof of the main result
By $b=-alpha>0$ we have
$$frac{1}{n^alpha cdot ln n }=frac{n^b}{ln n}=bfrac{n^b}{bln n}=bfrac{n^b}{ln n^b}$$
and since for any $b>0 implies n^bto infty$ we have that
$$bfrac{n^b}{ln n^b}to infty implies frac{1}{n^alpha cdot ln n } to infty$$
answered Dec 9 '18 at 15:39
gimusigimusi
93k84594
$endgroup$
add a comment |
$begingroup$
Preliminary results
We have that $forall a>0$
$$frac{e^{n}} {n^a} to infty$$
indeed by ratio test
$$frac{e^{n+1}}{(n+1)^a}frac {n^a} {e^n}to e>1$$
then from here we have that
$$frac{log n}{n} to 0 iff frac n {log n}to infty$$
indeed for any $a >0$
$$frac n {log n}ge a iff nge log n^a iff e^nge n^a$$
Proof of the main result
By $b=-alpha>0$ we have
$$frac{1}{n^alpha cdot ln n }=frac{n^b}{ln n}=bfrac{n^b}{bln n}=bfrac{n^b}{ln n^b}$$
and since for any $b>0 implies n^bto infty$ we have that
$$bfrac{n^b}{ln n^b}to infty implies frac{1}{n^alpha cdot ln n } to infty$$
answered Dec 9 '18 at 15:39
gimusigimusi
93k84594
$endgroup$
add a comment |
$begingroup$
Preliminary results
We have that $forall a>0$
$$frac{e^{n}} {n^a} to infty$$
indeed by ratio test
$$frac{e^{n+1}}{(n+1)^a}frac {n^a} {e^n}to e>1$$
then from here we have that
$$frac{log n}{n} to 0 iff frac n {log n}to infty$$
indeed for any $a >0$
$$frac n {log n}ge a iff nge log n^a iff e^nge n^a$$
Proof of the main result
By $b=-alpha>0$ we have
$$frac{1}{n^alpha cdot ln n }=frac{n^b}{ln n}=bfrac{n^b}{bln n}=bfrac{n^b}{ln n^b}$$
and since for any $b>0 implies n^bto infty$ we have that
$$bfrac{n^b}{ln n^b}to infty implies frac{1}{n^alpha cdot ln n } to infty$$
answered Dec 9 '18 at 15:39
gimusigimusi
93k84594
$endgroup$
Preliminary results
We have that $forall a>0$
$$frac{e^{n}} {n^a} to infty$$
indeed by ratio test
$$frac{e^{n+1}}{(n+1)^a}frac {n^a} {e^n}to e>1$$
then from here we have that
$$frac{log n}{n} to 0 iff frac n {log n}to infty$$
indeed for any $a >0$
$$frac n {log n}ge a iff nge log n^a iff e^nge n^a$$
Proof of the main result
By $b=-alpha>0$ we have
$$frac{1}{n^alpha cdot ln n }=frac{n^b}{ln n}=bfrac{n^b}{bln n}=bfrac{n^b}{ln n^b}$$
and since for any $b>0 implies n^bto infty$ we have that
$$bfrac{n^b}{ln n^b}to infty implies frac{1}{n^alpha cdot ln n } to infty$$
answered Dec 9 '18 at 15:39
gimusigimusi
93k84594
edited Dec 9 '18 at 19:17
answered Dec 9 '18 at 15:39
gimusigimusi
93k84594
answered Dec 9 '18 at 15:39
gimusigimusi
93k84594
answered Dec 9 '18 at 15:39
gimusigimusi
93k84594
93k84594
add a comment |
add a comment |
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$begingroup$
You are looking for a solution only by sequence? You should add that to your question in order to make it clear.
$endgroup$
– gimusi
Dec 9 '18 at 15:45
$begingroup$
@gimusi Okay. sorry. I edited the post.
$endgroup$
– Omer
Dec 9 '18 at 15:48
2
$begingroup$
What properties of the $log$ function do you know, yet?
$endgroup$
– Martin Rosenau
Dec 9 '18 at 16:13