Sequence problem regarding convergence from an online contest
$begingroup$
Let $(x_n)_{nin mathbb{N}}$ be a sequence defined by $x_0=1$ and $x_n=x_{n-1}cdot (1-frac{1}{4n^2})$,$forall ngeq 1$.
Prove that :
a)$(x_n)_{nin mathbb{N}}$ is convergent
b) if $l=lim_{nto infty} x_n$,compute $lim_{n to infty} (frac{x_n}{l})^n$.
What I did was substitute $n-1,n-2,....1$ in the recurrence relation and I got that $x_n=prod_{k=1}^{n} (1-frac{1}{4k^2})$.However,here I got stuck because I don't know how to find this limit.
real-analysis sequences-and-series
asked Dec 20 '18 at 16:38
Math GuyMath Guy
576
$endgroup$
|
show 2 more comments
$begingroup$
Let $(x_n)_{nin mathbb{N}}$ be a sequence defined by $x_0=1$ and $x_n=x_{n-1}cdot (1-frac{1}{4n^2})$,$forall ngeq 1$.
Prove that :
a)$(x_n)_{nin mathbb{N}}$ is convergent
b) if $l=lim_{nto infty} x_n$,compute $lim_{n to infty} (frac{x_n}{l})^n$.
What I did was substitute $n-1,n-2,....1$ in the recurrence relation and I got that $x_n=prod_{k=1}^{n} (1-frac{1}{4k^2})$.However,here I got stuck because I don't know how to find this limit.
real-analysis sequences-and-series
asked Dec 20 '18 at 16:38
Math GuyMath Guy
576
$endgroup$
$begingroup$
If it's for a contest, why should we help you rather than the other contestants (or enter the contest ourselves)?
$endgroup$
– Robert Israel
Dec 20 '18 at 16:49
$begingroup$
It is not from an active contest. The contest finished 4 days ago
$endgroup$
– Math Guy
Dec 20 '18 at 16:51
$begingroup$
disregard my answer, it is wrong
$endgroup$
– roman
Dec 20 '18 at 17:05
$begingroup$
See Wallis's product.
$endgroup$
– Robert Israel
Dec 20 '18 at 17:32
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@RobertIsrael thank you,I hadn't seen this result before. This means that $l=frac{2}{pi}$,hence $x_n$ is convergent.Could you also give me a hint for b)?I got that that limit is equal to $e^{lim_{ntoinfty}nleft(frac{x_{n}}{l}-1right)}$,but I don't know how to compute this.
$endgroup$
– Math Guy
Dec 20 '18 at 18:04
|
show 2 more comments
$begingroup$
Let $(x_n)_{nin mathbb{N}}$ be a sequence defined by $x_0=1$ and $x_n=x_{n-1}cdot (1-frac{1}{4n^2})$,$forall ngeq 1$.
Prove that :
a)$(x_n)_{nin mathbb{N}}$ is convergent
b) if $l=lim_{nto infty} x_n$,compute $lim_{n to infty} (frac{x_n}{l})^n$.
What I did was substitute $n-1,n-2,....1$ in the recurrence relation and I got that $x_n=prod_{k=1}^{n} (1-frac{1}{4k^2})$.However,here I got stuck because I don't know how to find this limit.
real-analysis sequences-and-series
asked Dec 20 '18 at 16:38
Math GuyMath Guy
576
$endgroup$
Let $(x_n)_{nin mathbb{N}}$ be a sequence defined by $x_0=1$ and $x_n=x_{n-1}cdot (1-frac{1}{4n^2})$,$forall ngeq 1$.
Prove that :
a)$(x_n)_{nin mathbb{N}}$ is convergent
b) if $l=lim_{nto infty} x_n$,compute $lim_{n to infty} (frac{x_n}{l})^n$.
What I did was substitute $n-1,n-2,....1$ in the recurrence relation and I got that $x_n=prod_{k=1}^{n} (1-frac{1}{4k^2})$.However,here I got stuck because I don't know how to find this limit.
real-analysis sequences-and-series
real-analysis sequences-and-series
asked Dec 20 '18 at 16:38
Math GuyMath Guy
576
asked Dec 20 '18 at 16:38
Math GuyMath Guy
576
asked Dec 20 '18 at 16:38
Math GuyMath Guy
576
asked Dec 20 '18 at 16:38
Math GuyMath Guy
576
asked Dec 20 '18 at 16:38
Math GuyMath Guy
576
576
$begingroup$
If it's for a contest, why should we help you rather than the other contestants (or enter the contest ourselves)?
$endgroup$
– Robert Israel
Dec 20 '18 at 16:49
$begingroup$
It is not from an active contest. The contest finished 4 days ago
$endgroup$
– Math Guy
Dec 20 '18 at 16:51
$begingroup$
disregard my answer, it is wrong
$endgroup$
– roman
Dec 20 '18 at 17:05
$begingroup$
See Wallis's product.
$endgroup$
– Robert Israel
Dec 20 '18 at 17:32
$begingroup$
@RobertIsrael thank you,I hadn't seen this result before. This means that $l=frac{2}{pi}$,hence $x_n$ is convergent.Could you also give me a hint for b)?I got that that limit is equal to $e^{lim_{ntoinfty}nleft(frac{x_{n}}{l}-1right)}$,but I don't know how to compute this.
$endgroup$
– Math Guy
Dec 20 '18 at 18:04
|
show 2 more comments
$begingroup$
If it's for a contest, why should we help you rather than the other contestants (or enter the contest ourselves)?
$endgroup$
– Robert Israel
Dec 20 '18 at 16:49
$begingroup$
It is not from an active contest. The contest finished 4 days ago
$endgroup$
– Math Guy
Dec 20 '18 at 16:51
$begingroup$
disregard my answer, it is wrong
$endgroup$
– roman
Dec 20 '18 at 17:05
$begingroup$
See Wallis's product.
$endgroup$
– Robert Israel
Dec 20 '18 at 17:32
$begingroup$
@RobertIsrael thank you,I hadn't seen this result before. This means that $l=frac{2}{pi}$,hence $x_n$ is convergent.Could you also give me a hint for b)?I got that that limit is equal to $e^{lim_{ntoinfty}nleft(frac{x_{n}}{l}-1right)}$,but I don't know how to compute this.
$endgroup$
– Math Guy
Dec 20 '18 at 18:04
$begingroup$
If it's for a contest, why should we help you rather than the other contestants (or enter the contest ourselves)?
$endgroup$
– Robert Israel
Dec 20 '18 at 16:49
$begingroup$
If it's for a contest, why should we help you rather than the other contestants (or enter the contest ourselves)?
$endgroup$
– Robert Israel
Dec 20 '18 at 16:49
$begingroup$
It is not from an active contest. The contest finished 4 days ago
$endgroup$
– Math Guy
Dec 20 '18 at 16:51
$begingroup$
It is not from an active contest. The contest finished 4 days ago
$endgroup$
– Math Guy
Dec 20 '18 at 16:51
$begingroup$
disregard my answer, it is wrong
$endgroup$
– roman
Dec 20 '18 at 17:05
$begingroup$
disregard my answer, it is wrong
$endgroup$
– roman
Dec 20 '18 at 17:05
$begingroup$
See Wallis's product.
$endgroup$
– Robert Israel
Dec 20 '18 at 17:32
$begingroup$
See Wallis's product.
$endgroup$
– Robert Israel
Dec 20 '18 at 17:32
$begingroup$
@RobertIsrael thank you,I hadn't seen this result before. This means that $l=frac{2}{pi}$,hence $x_n$ is convergent.Could you also give me a hint for b)?I got that that limit is equal to $e^{lim_{ntoinfty}nleft(frac{x_{n}}{l}-1right)}$,but I don't know how to compute this.
$endgroup$
– Math Guy
Dec 20 '18 at 18:04
$begingroup$
@RobertIsrael thank you,I hadn't seen this result before. This means that $l=frac{2}{pi}$,hence $x_n$ is convergent.Could you also give me a hint for b)?I got that that limit is equal to $e^{lim_{ntoinfty}nleft(frac{x_{n}}{l}-1right)}$,but I don't know how to compute this.
$endgroup$
– Math Guy
Dec 20 '18 at 18:04
|
show 2 more comments
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$begingroup$
If it's for a contest, why should we help you rather than the other contestants (or enter the contest ourselves)?
$endgroup$
– Robert Israel
Dec 20 '18 at 16:49
$begingroup$
It is not from an active contest. The contest finished 4 days ago
$endgroup$
– Math Guy
Dec 20 '18 at 16:51
$begingroup$
disregard my answer, it is wrong
$endgroup$
– roman
Dec 20 '18 at 17:05
$begingroup$
See Wallis's product.
$endgroup$
– Robert Israel
Dec 20 '18 at 17:32
$begingroup$
@RobertIsrael thank you,I hadn't seen this result before. This means that $l=frac{2}{pi}$,hence $x_n$ is convergent.Could you also give me a hint for b)?I got that that limit is equal to $e^{lim_{ntoinfty}nleft(frac{x_{n}}{l}-1right)}$,but I don't know how to compute this.
$endgroup$
– Math Guy
Dec 20 '18 at 18:04