Proof verification for $lim_{ntoinfty}frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}$
$begingroup$
Find the limit:
$$
lim_{ntoinfty}frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} text{for} a >0
$$
Let:
$$
begin{cases}
x_n = frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}\
ninmathbb N
end{cases}
$$
By definition of exponent we may rewrite $x_n$ as:
$$
x_n = frac{a}{1+a}cdotfrac{a}{1+a^2}cdotsfrac{a}{1+a^n}
$$
Since $a>0$:
$$
forall a>0,kinmathbb N: frac{a}{1+a^k} < 1
$$
Choose $q_n$ to be in the following form:
$$
q_n = maxleft{frac{a}{1+a}, frac{a}{1+a^2}, dots,frac{a}{1+a^n}right}
$$
It follows that:
$$
x_n = frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} le q^n_n
$$
Also $x_n > 0$. Now squeezing $x_n$ one may obtain:
$$
0 le lim_{ntoinfty} x_n le lim_{ntoinfty} q^n_n = 0
$$
Thus:
$$
lim_{ntoinfty}frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} = 0
$$
Have i missed something?
calculus limits proof-verification limits-without-lhopital
asked Dec 5 '18 at 15:46
romanroman
2,31121224
$endgroup$
add a comment |
$begingroup$
Find the limit:
$$
lim_{ntoinfty}frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} text{for} a >0
$$
Let:
$$
begin{cases}
x_n = frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}\
ninmathbb N
end{cases}
$$
By definition of exponent we may rewrite $x_n$ as:
$$
x_n = frac{a}{1+a}cdotfrac{a}{1+a^2}cdotsfrac{a}{1+a^n}
$$
Since $a>0$:
$$
forall a>0,kinmathbb N: frac{a}{1+a^k} < 1
$$
Choose $q_n$ to be in the following form:
$$
q_n = maxleft{frac{a}{1+a}, frac{a}{1+a^2}, dots,frac{a}{1+a^n}right}
$$
It follows that:
$$
x_n = frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} le q^n_n
$$
Also $x_n > 0$. Now squeezing $x_n$ one may obtain:
$$
0 le lim_{ntoinfty} x_n le lim_{ntoinfty} q^n_n = 0
$$
Thus:
$$
lim_{ntoinfty}frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} = 0
$$
Have i missed something?
calculus limits proof-verification limits-without-lhopital
asked Dec 5 '18 at 15:46
romanroman
2,31121224
$endgroup$
1
$begingroup$
The logic is fine for me. Two notation details: when you define a sequence (like your $x_n$) don't forget the scope of $n$ ('for all $ninBbb N'$, for example). And the equation after 'Since $a>0$' doesn't need any bracket.
$endgroup$
– ajotatxe
Dec 5 '18 at 15:52
2
$begingroup$
The $q$, defined as the maximum over $frac a{1+a},...,frac a{1+a^n}$ depends on $n$, therefore it should be written as $q_n$. Then, it is not clear whether $q_n^n$ goes to zero. This is an issue.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 15:53
1
$begingroup$
One more detail: $x_nle q_n$ (strict inequality may be false, and you don't need it).
$endgroup$
– ajotatxe
Dec 5 '18 at 15:54
$begingroup$
Thank you for the notices
$endgroup$
– roman
Dec 5 '18 at 15:57
add a comment |
$begingroup$
Find the limit:
$$
lim_{ntoinfty}frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} text{for} a >0
$$
Let:
$$
begin{cases}
x_n = frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}\
ninmathbb N
end{cases}
$$
By definition of exponent we may rewrite $x_n$ as:
$$
x_n = frac{a}{1+a}cdotfrac{a}{1+a^2}cdotsfrac{a}{1+a^n}
$$
Since $a>0$:
$$
forall a>0,kinmathbb N: frac{a}{1+a^k} < 1
$$
Choose $q_n$ to be in the following form:
$$
q_n = maxleft{frac{a}{1+a}, frac{a}{1+a^2}, dots,frac{a}{1+a^n}right}
$$
It follows that:
$$
x_n = frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} le q^n_n
$$
Also $x_n > 0$. Now squeezing $x_n$ one may obtain:
$$
0 le lim_{ntoinfty} x_n le lim_{ntoinfty} q^n_n = 0
$$
Thus:
$$
lim_{ntoinfty}frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} = 0
$$
Have i missed something?
calculus limits proof-verification limits-without-lhopital
asked Dec 5 '18 at 15:46
romanroman
2,31121224
$endgroup$
Find the limit:
$$
lim_{ntoinfty}frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} text{for} a >0
$$
Let:
$$
begin{cases}
x_n = frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}\
ninmathbb N
end{cases}
$$
By definition of exponent we may rewrite $x_n$ as:
$$
x_n = frac{a}{1+a}cdotfrac{a}{1+a^2}cdotsfrac{a}{1+a^n}
$$
Since $a>0$:
$$
forall a>0,kinmathbb N: frac{a}{1+a^k} < 1
$$
Choose $q_n$ to be in the following form:
$$
q_n = maxleft{frac{a}{1+a}, frac{a}{1+a^2}, dots,frac{a}{1+a^n}right}
$$
It follows that:
$$
x_n = frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} le q^n_n
$$
Also $x_n > 0$. Now squeezing $x_n$ one may obtain:
$$
0 le lim_{ntoinfty} x_n le lim_{ntoinfty} q^n_n = 0
$$
Thus:
$$
lim_{ntoinfty}frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} = 0
$$
Have i missed something?
calculus limits proof-verification limits-without-lhopital
calculus limits proof-verification limits-without-lhopital
asked Dec 5 '18 at 15:46
romanroman
2,31121224
asked Dec 5 '18 at 15:46
romanroman
2,31121224
edited Dec 5 '18 at 15:57
roman
asked Dec 5 '18 at 15:46
romanroman
2,31121224
asked Dec 5 '18 at 15:46
romanroman
2,31121224
asked Dec 5 '18 at 15:46
romanroman
2,31121224
2,31121224
1
$begingroup$
The logic is fine for me. Two notation details: when you define a sequence (like your $x_n$) don't forget the scope of $n$ ('for all $ninBbb N'$, for example). And the equation after 'Since $a>0$' doesn't need any bracket.
$endgroup$
– ajotatxe
Dec 5 '18 at 15:52
2
$begingroup$
The $q$, defined as the maximum over $frac a{1+a},...,frac a{1+a^n}$ depends on $n$, therefore it should be written as $q_n$. Then, it is not clear whether $q_n^n$ goes to zero. This is an issue.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 15:53
1
$begingroup$
One more detail: $x_nle q_n$ (strict inequality may be false, and you don't need it).
$endgroup$
– ajotatxe
Dec 5 '18 at 15:54
$begingroup$
Thank you for the notices
$endgroup$
– roman
Dec 5 '18 at 15:57
add a comment |
1
$begingroup$
The logic is fine for me. Two notation details: when you define a sequence (like your $x_n$) don't forget the scope of $n$ ('for all $ninBbb N'$, for example). And the equation after 'Since $a>0$' doesn't need any bracket.
$endgroup$
– ajotatxe
Dec 5 '18 at 15:52
2
$begingroup$
The $q$, defined as the maximum over $frac a{1+a},...,frac a{1+a^n}$ depends on $n$, therefore it should be written as $q_n$. Then, it is not clear whether $q_n^n$ goes to zero. This is an issue.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 15:53
1
$begingroup$
One more detail: $x_nle q_n$ (strict inequality may be false, and you don't need it).
$endgroup$
– ajotatxe
Dec 5 '18 at 15:54
$begingroup$
Thank you for the notices
$endgroup$
– roman
Dec 5 '18 at 15:57
1
1
$begingroup$
The logic is fine for me. Two notation details: when you define a sequence (like your $x_n$) don't forget the scope of $n$ ('for all $ninBbb N'$, for example). And the equation after 'Since $a>0$' doesn't need any bracket.
$endgroup$
– ajotatxe
Dec 5 '18 at 15:52
$begingroup$
The logic is fine for me. Two notation details: when you define a sequence (like your $x_n$) don't forget the scope of $n$ ('for all $ninBbb N'$, for example). And the equation after 'Since $a>0$' doesn't need any bracket.
$endgroup$
– ajotatxe
Dec 5 '18 at 15:52
2
2
$begingroup$
The $q$, defined as the maximum over $frac a{1+a},...,frac a{1+a^n}$ depends on $n$, therefore it should be written as $q_n$. Then, it is not clear whether $q_n^n$ goes to zero. This is an issue.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 15:53
$begingroup$
The $q$, defined as the maximum over $frac a{1+a},...,frac a{1+a^n}$ depends on $n$, therefore it should be written as $q_n$. Then, it is not clear whether $q_n^n$ goes to zero. This is an issue.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 15:53
1
1
$begingroup$
One more detail: $x_nle q_n$ (strict inequality may be false, and you don't need it).
$endgroup$
– ajotatxe
Dec 5 '18 at 15:54
$begingroup$
One more detail: $x_nle q_n$ (strict inequality may be false, and you don't need it).
$endgroup$
– ajotatxe
Dec 5 '18 at 15:54
$begingroup$
Thank you for the notices
$endgroup$
– roman
Dec 5 '18 at 15:57
$begingroup$
Thank you for the notices
$endgroup$
– roman
Dec 5 '18 at 15:57
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
HINT
As an alternative by ratio test
$$frac{x_{n+1}}{x_n}=frac{a^{n+1}}{(1+a)(1+a^2)cdots(1+a^{n+1})}frac{(1+a)(1+a^2)cdots(1+a^n)}{a^n}=frac{a}{1+a^{n+1}}$$
answered Dec 5 '18 at 15:57
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
I find this approach much more elegant, thanks!
$endgroup$
– roman
Dec 5 '18 at 15:58
$begingroup$
@roman Yes the good old ratio test seems very effective here :)
$endgroup$
– gimusi
Dec 5 '18 at 15:59
2
$begingroup$
Nice application of the ratio test (+1).
$endgroup$
– Robert Z
Dec 5 '18 at 16:03
$begingroup$
@RobertZ Thanks a lot much appreciative from you! Regards
$endgroup$
– gimusi
Dec 5 '18 at 16:05
add a comment |
$begingroup$
Following your idea to apply the squeeze theorem, the proof becomes simpler if we note that $0leq a<1$,
$$0leq frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}leq a^nto 0$$
and for $ageq 1$,
$$0leq frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}leq frac{a^n}{2^n a^{n(n+1)/4}}to 0$$
where we used the fact that $1+a^kgeq 2a^{k/2}.$
answered Dec 5 '18 at 16:00
Robert ZRobert Z
99.1k1068139
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT
As an alternative by ratio test
$$frac{x_{n+1}}{x_n}=frac{a^{n+1}}{(1+a)(1+a^2)cdots(1+a^{n+1})}frac{(1+a)(1+a^2)cdots(1+a^n)}{a^n}=frac{a}{1+a^{n+1}}$$
answered Dec 5 '18 at 15:57
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
I find this approach much more elegant, thanks!
$endgroup$
– roman
Dec 5 '18 at 15:58
$begingroup$
@roman Yes the good old ratio test seems very effective here :)
$endgroup$
– gimusi
Dec 5 '18 at 15:59
2
$begingroup$
Nice application of the ratio test (+1).
$endgroup$
– Robert Z
Dec 5 '18 at 16:03
$begingroup$
@RobertZ Thanks a lot much appreciative from you! Regards
$endgroup$
– gimusi
Dec 5 '18 at 16:05
add a comment |
$begingroup$
HINT
As an alternative by ratio test
$$frac{x_{n+1}}{x_n}=frac{a^{n+1}}{(1+a)(1+a^2)cdots(1+a^{n+1})}frac{(1+a)(1+a^2)cdots(1+a^n)}{a^n}=frac{a}{1+a^{n+1}}$$
answered Dec 5 '18 at 15:57
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
I find this approach much more elegant, thanks!
$endgroup$
– roman
Dec 5 '18 at 15:58
$begingroup$
@roman Yes the good old ratio test seems very effective here :)
$endgroup$
– gimusi
Dec 5 '18 at 15:59
2
$begingroup$
Nice application of the ratio test (+1).
$endgroup$
– Robert Z
Dec 5 '18 at 16:03
$begingroup$
@RobertZ Thanks a lot much appreciative from you! Regards
$endgroup$
– gimusi
Dec 5 '18 at 16:05
add a comment |
$begingroup$
HINT
As an alternative by ratio test
$$frac{x_{n+1}}{x_n}=frac{a^{n+1}}{(1+a)(1+a^2)cdots(1+a^{n+1})}frac{(1+a)(1+a^2)cdots(1+a^n)}{a^n}=frac{a}{1+a^{n+1}}$$
answered Dec 5 '18 at 15:57
gimusigimusi
92.9k84494
$endgroup$
HINT
As an alternative by ratio test
$$frac{x_{n+1}}{x_n}=frac{a^{n+1}}{(1+a)(1+a^2)cdots(1+a^{n+1})}frac{(1+a)(1+a^2)cdots(1+a^n)}{a^n}=frac{a}{1+a^{n+1}}$$
answered Dec 5 '18 at 15:57
gimusigimusi
92.9k84494
answered Dec 5 '18 at 15:57
gimusigimusi
92.9k84494
answered Dec 5 '18 at 15:57
gimusigimusi
92.9k84494
answered Dec 5 '18 at 15:57
gimusigimusi
92.9k84494
92.9k84494
$begingroup$
I find this approach much more elegant, thanks!
$endgroup$
– roman
Dec 5 '18 at 15:58
$begingroup$
@roman Yes the good old ratio test seems very effective here :)
$endgroup$
– gimusi
Dec 5 '18 at 15:59
2
$begingroup$
Nice application of the ratio test (+1).
$endgroup$
– Robert Z
Dec 5 '18 at 16:03
$begingroup$
@RobertZ Thanks a lot much appreciative from you! Regards
$endgroup$
– gimusi
Dec 5 '18 at 16:05
add a comment |
$begingroup$
I find this approach much more elegant, thanks!
$endgroup$
– roman
Dec 5 '18 at 15:58
$begingroup$
@roman Yes the good old ratio test seems very effective here :)
$endgroup$
– gimusi
Dec 5 '18 at 15:59
2
$begingroup$
Nice application of the ratio test (+1).
$endgroup$
– Robert Z
Dec 5 '18 at 16:03
$begingroup$
@RobertZ Thanks a lot much appreciative from you! Regards
$endgroup$
– gimusi
Dec 5 '18 at 16:05
$begingroup$
I find this approach much more elegant, thanks!
$endgroup$
– roman
Dec 5 '18 at 15:58
$begingroup$
I find this approach much more elegant, thanks!
$endgroup$
– roman
Dec 5 '18 at 15:58
$begingroup$
@roman Yes the good old ratio test seems very effective here :)
$endgroup$
– gimusi
Dec 5 '18 at 15:59
$begingroup$
@roman Yes the good old ratio test seems very effective here :)
$endgroup$
– gimusi
Dec 5 '18 at 15:59
2
2
$begingroup$
Nice application of the ratio test (+1).
$endgroup$
– Robert Z
Dec 5 '18 at 16:03
$begingroup$
Nice application of the ratio test (+1).
$endgroup$
– Robert Z
Dec 5 '18 at 16:03
$begingroup$
@RobertZ Thanks a lot much appreciative from you! Regards
$endgroup$
– gimusi
Dec 5 '18 at 16:05
$begingroup$
@RobertZ Thanks a lot much appreciative from you! Regards
$endgroup$
– gimusi
Dec 5 '18 at 16:05
add a comment |
$begingroup$
Following your idea to apply the squeeze theorem, the proof becomes simpler if we note that $0leq a<1$,
$$0leq frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}leq a^nto 0$$
and for $ageq 1$,
$$0leq frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}leq frac{a^n}{2^n a^{n(n+1)/4}}to 0$$
where we used the fact that $1+a^kgeq 2a^{k/2}.$
answered Dec 5 '18 at 16:00
Robert ZRobert Z
99.1k1068139
$endgroup$
add a comment |
$begingroup$
Following your idea to apply the squeeze theorem, the proof becomes simpler if we note that $0leq a<1$,
$$0leq frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}leq a^nto 0$$
and for $ageq 1$,
$$0leq frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}leq frac{a^n}{2^n a^{n(n+1)/4}}to 0$$
where we used the fact that $1+a^kgeq 2a^{k/2}.$
answered Dec 5 '18 at 16:00
Robert ZRobert Z
99.1k1068139
$endgroup$
add a comment |
$begingroup$
Following your idea to apply the squeeze theorem, the proof becomes simpler if we note that $0leq a<1$,
$$0leq frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}leq a^nto 0$$
and for $ageq 1$,
$$0leq frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}leq frac{a^n}{2^n a^{n(n+1)/4}}to 0$$
where we used the fact that $1+a^kgeq 2a^{k/2}.$
answered Dec 5 '18 at 16:00
Robert ZRobert Z
99.1k1068139
$endgroup$
Following your idea to apply the squeeze theorem, the proof becomes simpler if we note that $0leq a<1$,
$$0leq frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}leq a^nto 0$$
and for $ageq 1$,
$$0leq frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}leq frac{a^n}{2^n a^{n(n+1)/4}}to 0$$
where we used the fact that $1+a^kgeq 2a^{k/2}.$
answered Dec 5 '18 at 16:00
Robert ZRobert Z
99.1k1068139
edited Dec 5 '18 at 16:05
answered Dec 5 '18 at 16:00
Robert ZRobert Z
99.1k1068139
answered Dec 5 '18 at 16:00
Robert ZRobert Z
99.1k1068139
answered Dec 5 '18 at 16:00
Robert ZRobert Z
99.1k1068139
99.1k1068139
add a comment |
add a comment |
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1
$begingroup$
The logic is fine for me. Two notation details: when you define a sequence (like your $x_n$) don't forget the scope of $n$ ('for all $ninBbb N'$, for example). And the equation after 'Since $a>0$' doesn't need any bracket.
$endgroup$
– ajotatxe
Dec 5 '18 at 15:52
2
$begingroup$
The $q$, defined as the maximum over $frac a{1+a},...,frac a{1+a^n}$ depends on $n$, therefore it should be written as $q_n$. Then, it is not clear whether $q_n^n$ goes to zero. This is an issue.
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– астон вілла олоф мэллбэрг
Dec 5 '18 at 15:53
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One more detail: $x_nle q_n$ (strict inequality may be false, and you don't need it).
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– ajotatxe
Dec 5 '18 at 15:54
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Thank you for the notices
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– roman
Dec 5 '18 at 15:57