if $|a|<1$ so $lim_{nto infty}na^n=0$.
$begingroup$
Prove that if $|a|<1$ ($a$ is real) so $lim_{nto infty}na^n=0$.
I know that I need to use squeeze theory (because I have $-1<a<1$) but I dont see how.
thanks
sequences-and-series limits
asked Apr 5 '15 at 12:03
StabiloStabilo
734512
$endgroup$
add a comment |
$begingroup$
Prove that if $|a|<1$ ($a$ is real) so $lim_{nto infty}na^n=0$.
I know that I need to use squeeze theory (because I have $-1<a<1$) but I dont see how.
thanks
sequences-and-series limits
asked Apr 5 '15 at 12:03
StabiloStabilo
734512
$endgroup$
add a comment |
$begingroup$
Prove that if $|a|<1$ ($a$ is real) so $lim_{nto infty}na^n=0$.
I know that I need to use squeeze theory (because I have $-1<a<1$) but I dont see how.
thanks
sequences-and-series limits
asked Apr 5 '15 at 12:03
StabiloStabilo
734512
$endgroup$
Prove that if $|a|<1$ ($a$ is real) so $lim_{nto infty}na^n=0$.
I know that I need to use squeeze theory (because I have $-1<a<1$) but I dont see how.
thanks
sequences-and-series limits
sequences-and-series limits
asked Apr 5 '15 at 12:03
StabiloStabilo
734512
asked Apr 5 '15 at 12:03
StabiloStabilo
734512
asked Apr 5 '15 at 12:03
StabiloStabilo
734512
asked Apr 5 '15 at 12:03
StabiloStabilo
734512
asked Apr 5 '15 at 12:03
StabiloStabilo
734512
734512
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
If you want to do it by hand: since $|a| < 1$, there exists $N$ such that $frac{n+1}{n}|a| < q < 1, forall n geq N$. Hence $n|a|^n = frac{n}{n-1}|a|frac{n-1}{n-2}|a|dotsfrac{N+1}{N}|a| cdot N|a|^N < q^{n-N} cdot N|a|^N, , forall ngeq N.$
The latter converges to zero and you get your squeezing sequence(s).
answered Apr 5 '15 at 12:19
user130065user130065
1362
$endgroup$
$begingroup$
This is, as asked by the OP, the simplest, most basic answer here imo, with only basic stuff about sequences. +1
$endgroup$
– Timbuc
Apr 5 '15 at 12:31
add a comment |
$begingroup$
If $|a| < 1,$ we claim that the series $sum_{1}^{infty}na^{n}$ converges, so that
$na^{n} to 0$. It suffices to show that the series is absolutely convergent. But, since
$$frac{(n+1)|a|^{n+1}}{n|a|^{n}} = (1+frac{1}{n})|a| to |a| < 1,$$ by the ratio test the series converges. Thus we have
$$lim_{n to infty}na^{n} = 0.$$
answered Apr 5 '15 at 12:28
Gary MooreGary Moore
17.3k21546
$endgroup$
add a comment |
$begingroup$
You can use the relationship of sequences limits and function limits, just let $0<a<1$ for convenient, and study $xa^x,xrightarrowinfty$ you can use L.Hospital theorem, $lim frac{x}{a^{-x}}=lim frac{1}{-a^{-x}}frac{1}{lna}=0 $
answered Apr 5 '15 at 12:17
HansHans
12210
$endgroup$
$begingroup$
just sequences, thanks
$endgroup$
– Stabilo
Apr 5 '15 at 12:19
$begingroup$
Oh, that's ok... in other conditions where you can use L.Hospital, don't forget it and it's so powerful.
$endgroup$
– Hans
Apr 5 '15 at 12:23
add a comment |
$begingroup$
Here a very elementary way:
$|a| < 1 stackrel{a neq 0}{Rightarrow} |a| = frac{1}{1+p}$ for a $p > 0$.- $(1+p)^n = 1 + np + color{blue}{frac{n(n-1)}{2}p^2} + cdots + p^n color{blue}{>frac{n(n-1)}{2}p^2}$
So, it follows
$$0 leq |na^n| = n|a|^n = frac{n}{(1+p)^n} < frac{n}{color{blue}{frac{n(n-1)}{2}p^2}} = frac{2}{(n-1)p^2} stackrel{n to infty}{longrightarrow} 0$$
answered Dec 9 '18 at 17:35
trancelocationtrancelocation
12.9k1827
$endgroup$
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
$begingroup$
If you want to do it by hand: since $|a| < 1$, there exists $N$ such that $frac{n+1}{n}|a| < q < 1, forall n geq N$. Hence $n|a|^n = frac{n}{n-1}|a|frac{n-1}{n-2}|a|dotsfrac{N+1}{N}|a| cdot N|a|^N < q^{n-N} cdot N|a|^N, , forall ngeq N.$
The latter converges to zero and you get your squeezing sequence(s).
answered Apr 5 '15 at 12:19
user130065user130065
1362
$endgroup$
$begingroup$
This is, as asked by the OP, the simplest, most basic answer here imo, with only basic stuff about sequences. +1
$endgroup$
– Timbuc
Apr 5 '15 at 12:31
add a comment |
$begingroup$
If you want to do it by hand: since $|a| < 1$, there exists $N$ such that $frac{n+1}{n}|a| < q < 1, forall n geq N$. Hence $n|a|^n = frac{n}{n-1}|a|frac{n-1}{n-2}|a|dotsfrac{N+1}{N}|a| cdot N|a|^N < q^{n-N} cdot N|a|^N, , forall ngeq N.$
The latter converges to zero and you get your squeezing sequence(s).
answered Apr 5 '15 at 12:19
user130065user130065
1362
$endgroup$
$begingroup$
This is, as asked by the OP, the simplest, most basic answer here imo, with only basic stuff about sequences. +1
$endgroup$
– Timbuc
Apr 5 '15 at 12:31
add a comment |
$begingroup$
If you want to do it by hand: since $|a| < 1$, there exists $N$ such that $frac{n+1}{n}|a| < q < 1, forall n geq N$. Hence $n|a|^n = frac{n}{n-1}|a|frac{n-1}{n-2}|a|dotsfrac{N+1}{N}|a| cdot N|a|^N < q^{n-N} cdot N|a|^N, , forall ngeq N.$
The latter converges to zero and you get your squeezing sequence(s).
answered Apr 5 '15 at 12:19
user130065user130065
1362
$endgroup$
If you want to do it by hand: since $|a| < 1$, there exists $N$ such that $frac{n+1}{n}|a| < q < 1, forall n geq N$. Hence $n|a|^n = frac{n}{n-1}|a|frac{n-1}{n-2}|a|dotsfrac{N+1}{N}|a| cdot N|a|^N < q^{n-N} cdot N|a|^N, , forall ngeq N.$
The latter converges to zero and you get your squeezing sequence(s).
answered Apr 5 '15 at 12:19
user130065user130065
1362
answered Apr 5 '15 at 12:19
user130065user130065
1362
answered Apr 5 '15 at 12:19
user130065user130065
1362
answered Apr 5 '15 at 12:19
user130065user130065
1362
1362
$begingroup$
This is, as asked by the OP, the simplest, most basic answer here imo, with only basic stuff about sequences. +1
$endgroup$
– Timbuc
Apr 5 '15 at 12:31
add a comment |
$begingroup$
This is, as asked by the OP, the simplest, most basic answer here imo, with only basic stuff about sequences. +1
$endgroup$
– Timbuc
Apr 5 '15 at 12:31
$begingroup$
This is, as asked by the OP, the simplest, most basic answer here imo, with only basic stuff about sequences. +1
$endgroup$
– Timbuc
Apr 5 '15 at 12:31
$begingroup$
This is, as asked by the OP, the simplest, most basic answer here imo, with only basic stuff about sequences. +1
$endgroup$
– Timbuc
Apr 5 '15 at 12:31
add a comment |
$begingroup$
If $|a| < 1,$ we claim that the series $sum_{1}^{infty}na^{n}$ converges, so that
$na^{n} to 0$. It suffices to show that the series is absolutely convergent. But, since
$$frac{(n+1)|a|^{n+1}}{n|a|^{n}} = (1+frac{1}{n})|a| to |a| < 1,$$ by the ratio test the series converges. Thus we have
$$lim_{n to infty}na^{n} = 0.$$
answered Apr 5 '15 at 12:28
Gary MooreGary Moore
17.3k21546
$endgroup$
add a comment |
$begingroup$
If $|a| < 1,$ we claim that the series $sum_{1}^{infty}na^{n}$ converges, so that
$na^{n} to 0$. It suffices to show that the series is absolutely convergent. But, since
$$frac{(n+1)|a|^{n+1}}{n|a|^{n}} = (1+frac{1}{n})|a| to |a| < 1,$$ by the ratio test the series converges. Thus we have
$$lim_{n to infty}na^{n} = 0.$$
answered Apr 5 '15 at 12:28
Gary MooreGary Moore
17.3k21546
$endgroup$
add a comment |
$begingroup$
If $|a| < 1,$ we claim that the series $sum_{1}^{infty}na^{n}$ converges, so that
$na^{n} to 0$. It suffices to show that the series is absolutely convergent. But, since
$$frac{(n+1)|a|^{n+1}}{n|a|^{n}} = (1+frac{1}{n})|a| to |a| < 1,$$ by the ratio test the series converges. Thus we have
$$lim_{n to infty}na^{n} = 0.$$
answered Apr 5 '15 at 12:28
Gary MooreGary Moore
17.3k21546
$endgroup$
If $|a| < 1,$ we claim that the series $sum_{1}^{infty}na^{n}$ converges, so that
$na^{n} to 0$. It suffices to show that the series is absolutely convergent. But, since
$$frac{(n+1)|a|^{n+1}}{n|a|^{n}} = (1+frac{1}{n})|a| to |a| < 1,$$ by the ratio test the series converges. Thus we have
$$lim_{n to infty}na^{n} = 0.$$
answered Apr 5 '15 at 12:28
Gary MooreGary Moore
17.3k21546
answered Apr 5 '15 at 12:28
Gary MooreGary Moore
17.3k21546
answered Apr 5 '15 at 12:28
Gary MooreGary Moore
17.3k21546
answered Apr 5 '15 at 12:28
Gary MooreGary Moore
17.3k21546
17.3k21546
add a comment |
add a comment |
$begingroup$
You can use the relationship of sequences limits and function limits, just let $0<a<1$ for convenient, and study $xa^x,xrightarrowinfty$ you can use L.Hospital theorem, $lim frac{x}{a^{-x}}=lim frac{1}{-a^{-x}}frac{1}{lna}=0 $
answered Apr 5 '15 at 12:17
HansHans
12210
$endgroup$
$begingroup$
just sequences, thanks
$endgroup$
– Stabilo
Apr 5 '15 at 12:19
$begingroup$
Oh, that's ok... in other conditions where you can use L.Hospital, don't forget it and it's so powerful.
$endgroup$
– Hans
Apr 5 '15 at 12:23
add a comment |
$begingroup$
You can use the relationship of sequences limits and function limits, just let $0<a<1$ for convenient, and study $xa^x,xrightarrowinfty$ you can use L.Hospital theorem, $lim frac{x}{a^{-x}}=lim frac{1}{-a^{-x}}frac{1}{lna}=0 $
answered Apr 5 '15 at 12:17
HansHans
12210
$endgroup$
$begingroup$
just sequences, thanks
$endgroup$
– Stabilo
Apr 5 '15 at 12:19
$begingroup$
Oh, that's ok... in other conditions where you can use L.Hospital, don't forget it and it's so powerful.
$endgroup$
– Hans
Apr 5 '15 at 12:23
add a comment |
$begingroup$
You can use the relationship of sequences limits and function limits, just let $0<a<1$ for convenient, and study $xa^x,xrightarrowinfty$ you can use L.Hospital theorem, $lim frac{x}{a^{-x}}=lim frac{1}{-a^{-x}}frac{1}{lna}=0 $
answered Apr 5 '15 at 12:17
HansHans
12210
$endgroup$
You can use the relationship of sequences limits and function limits, just let $0<a<1$ for convenient, and study $xa^x,xrightarrowinfty$ you can use L.Hospital theorem, $lim frac{x}{a^{-x}}=lim frac{1}{-a^{-x}}frac{1}{lna}=0 $
answered Apr 5 '15 at 12:17
HansHans
12210
answered Apr 5 '15 at 12:17
HansHans
12210
answered Apr 5 '15 at 12:17
HansHans
12210
answered Apr 5 '15 at 12:17
HansHans
12210
12210
$begingroup$
just sequences, thanks
$endgroup$
– Stabilo
Apr 5 '15 at 12:19
$begingroup$
Oh, that's ok... in other conditions where you can use L.Hospital, don't forget it and it's so powerful.
$endgroup$
– Hans
Apr 5 '15 at 12:23
add a comment |
$begingroup$
just sequences, thanks
$endgroup$
– Stabilo
Apr 5 '15 at 12:19
$begingroup$
Oh, that's ok... in other conditions where you can use L.Hospital, don't forget it and it's so powerful.
$endgroup$
– Hans
Apr 5 '15 at 12:23
$begingroup$
just sequences, thanks
$endgroup$
– Stabilo
Apr 5 '15 at 12:19
$begingroup$
just sequences, thanks
$endgroup$
– Stabilo
Apr 5 '15 at 12:19
$begingroup$
Oh, that's ok... in other conditions where you can use L.Hospital, don't forget it and it's so powerful.
$endgroup$
– Hans
Apr 5 '15 at 12:23
$begingroup$
Oh, that's ok... in other conditions where you can use L.Hospital, don't forget it and it's so powerful.
$endgroup$
– Hans
Apr 5 '15 at 12:23
add a comment |
$begingroup$
Here a very elementary way:
$|a| < 1 stackrel{a neq 0}{Rightarrow} |a| = frac{1}{1+p}$ for a $p > 0$.- $(1+p)^n = 1 + np + color{blue}{frac{n(n-1)}{2}p^2} + cdots + p^n color{blue}{>frac{n(n-1)}{2}p^2}$
So, it follows
$$0 leq |na^n| = n|a|^n = frac{n}{(1+p)^n} < frac{n}{color{blue}{frac{n(n-1)}{2}p^2}} = frac{2}{(n-1)p^2} stackrel{n to infty}{longrightarrow} 0$$
answered Dec 9 '18 at 17:35
trancelocationtrancelocation
12.9k1827
$endgroup$
add a comment |
$begingroup$
Here a very elementary way:
$|a| < 1 stackrel{a neq 0}{Rightarrow} |a| = frac{1}{1+p}$ for a $p > 0$.- $(1+p)^n = 1 + np + color{blue}{frac{n(n-1)}{2}p^2} + cdots + p^n color{blue}{>frac{n(n-1)}{2}p^2}$
So, it follows
$$0 leq |na^n| = n|a|^n = frac{n}{(1+p)^n} < frac{n}{color{blue}{frac{n(n-1)}{2}p^2}} = frac{2}{(n-1)p^2} stackrel{n to infty}{longrightarrow} 0$$
answered Dec 9 '18 at 17:35
trancelocationtrancelocation
12.9k1827
$endgroup$
add a comment |
$begingroup$
Here a very elementary way:
$|a| < 1 stackrel{a neq 0}{Rightarrow} |a| = frac{1}{1+p}$ for a $p > 0$.- $(1+p)^n = 1 + np + color{blue}{frac{n(n-1)}{2}p^2} + cdots + p^n color{blue}{>frac{n(n-1)}{2}p^2}$
So, it follows
$$0 leq |na^n| = n|a|^n = frac{n}{(1+p)^n} < frac{n}{color{blue}{frac{n(n-1)}{2}p^2}} = frac{2}{(n-1)p^2} stackrel{n to infty}{longrightarrow} 0$$
answered Dec 9 '18 at 17:35
trancelocationtrancelocation
12.9k1827
$endgroup$
Here a very elementary way:
$|a| < 1 stackrel{a neq 0}{Rightarrow} |a| = frac{1}{1+p}$ for a $p > 0$.- $(1+p)^n = 1 + np + color{blue}{frac{n(n-1)}{2}p^2} + cdots + p^n color{blue}{>frac{n(n-1)}{2}p^2}$
So, it follows
$$0 leq |na^n| = n|a|^n = frac{n}{(1+p)^n} < frac{n}{color{blue}{frac{n(n-1)}{2}p^2}} = frac{2}{(n-1)p^2} stackrel{n to infty}{longrightarrow} 0$$
answered Dec 9 '18 at 17:35
trancelocationtrancelocation
12.9k1827
answered Dec 9 '18 at 17:35
trancelocationtrancelocation
12.9k1827
answered Dec 9 '18 at 17:35
trancelocationtrancelocation
12.9k1827
answered Dec 9 '18 at 17:35
trancelocationtrancelocation
12.9k1827
12.9k1827
add a comment |
add a comment |
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