Limit of a recursive sequence $s_n = (1-\frac{1}{4n^2})s_{n-1}$ I have $$s_1 = 1, s_n = (1-\frac{1}{4n^2})s_{n-1}.$$
I see that its limit exists, but cannot figure out what the limit is. How do I find its limit?
The only way I used (know) is:
$$\lim s_n = \lim s_{n+1}$$
But here, I am just getting $L = L$.
Also, when I program this sequence on computer it's approaching to $0.636619...$ and Wolfram Alpha is giving me complicated recurrence equation solution.
 A: Clearly, the limit of the sequence is equal to $\displaystyle \prod_{n = 2}^{\infty} \left(\frac{4n^2 - 1}{4n^2} \right)$. This is the reciprocal of the Wallis product (without the first term). The Wallis Product is equal to $\frac{\pi}{2}$ with the first term, $\frac{4}{3}$, so without it it is equal to $\frac{3 \pi}{8}$. Since this is the reciprocal, the desired answer is $\frac{8}{3 \pi}$. (By your program approximation, I suspect you meant to say that $s_0 = 1$ and not $s_1 = 1$, in which case the answer would just be $\frac{2}{\pi}$
One may prove the Wallis Product in many ways, originally by expressing $\frac{\sin x}{x}$ as a polynomial with roots at $\pm n \pi$ for $n \in \mathbb{Z}^+$. You may read about it here: http://en.wikipedia.org/wiki/Wallis_product
A: $\newcommand{\+}{^{\dagger}}
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$\ds{s_{1} = 1\,,\qquad s_{n} = \pars{1 - {1 \over 4n^{2}}}s_{n - 1}\,,\quad n \geq 2
     .\qquad\lim_{n \to \infty}s_{n}: {\large ?}}$

$$
s_{n} =\pars{1 - {1 \over 4n^{2}}}\bracks{1 - {1 \over 4\pars{n - 1}^{2}}}
\ldots \pars{1 - {1 \over 4\times 2^{2}}}s_{1}
$$

\begin{align}
\ln\pars{s_{n}}&=\overbrace{\ln\pars{s_{1}}}^{\ds{0}}\
+\ \sum_{k = 2}^{n}\ln\pars{1 - {1 \over 4k^{2}}}
=-\sum_{k = 2}^{n}\int_{0}^{1}{\dd x \over x + 4k^{2} - 1}
\\[3mm]&=-\,{1 \over 4}\int_{0}^{1}\sum_{k = 0}^{n - 2}
{1 \over \pars{k + 2 + \root{1 - x}/2}\pars{k + 2 - \root{1 - x}/2}}
\end{align}

With $\ds{\quad t \equiv \root{1 - x}\quad\imp\quad x = 1 - t^{2}}$:
  \begin{align}
\color{#c00000}{\lim_{n \to \infty}\ln\pars{s_{n}}}&
=-\,{1 \over 4}
\int_{0}^{1}{\Psi\pars{2 + \root{1 - x}/2} - \Psi\pars{2 - \root{1 - x}/2}
\over \root{1 - x}}\,\dd x
\\[3mm]&=-\,\half\int_{0}^{1}\bracks{%
\Psi\pars{2 + {t \over 2}} - \Psi\pars{2 - {t \over 2}}}\,\dd t
=-\left.\ln\pars{\Gamma\pars{2 + {t \over 2}}\Gamma\pars{2 - {t \over 2}}}
\vphantom{\LARGE A}\right\vert_{0}^{1}
\\[3mm]&=-\ln\pars{\Gamma\pars{5 \over 2}\Gamma\pars{3 \over 2}} +\ln\pars{\Gamma\pars{2}\Gamma\pars{2}}
=-\ln\pars{{3 \over 2}\,\Gamma^{2}\pars{3 \over 2}}
\\[3mm]&=-\ln\pars{{3 \over 2}\,{1 \over 4}\,\Gamma^{2}\pars{\half}}
=-\ln\pars{3\pi \over 8}=\color{#c00000}{\ln\pars{8 \over 3\pi}}
\end{align}

Then,
$$
\color{#00f}{\large\lim_{n \to \infty}s_{n} = {8 \over 3\pi}} \approx 0.8488
$$

$\ds{\Gamma\pars{z}}$ and $\ds{\Psi\pars{z}}$ are the Gamma and Digamma Functions, respectively.

A: Assume that $s_0=1$, then
$$
s_n=\frac{((2n)!)^2\,(2n+1)}{2^{4n}\,(n!)^4}.
$$
Stirling equivalent
$$
k!\sim\sqrt{2\pi k}\,k^k\,\mathrm e^{-k},
$$
then yields
$$
\lim\limits_{n\to\infty}s_n=\frac2\pi\approx0.63662.
$$
If $s_1=1$ instead of $s_1=\frac34$, then $$\lim\limits_{n\to\infty}s_n=\frac43\cdot\frac2\pi.$$
