I saw this problem:
Find $\lim\limits_{n \to \infty }\prod\limits_{r=1}^n \frac{n^2-r}{n^2+r} $
I tried to prove this problem and got:
$$\ln(1+x) = \sum_{k=1}^ \infty \frac{(-1)^{k+1} x^k}{ k} \ \ \ \ \text{ with radius of convergence = 1 }$$
$$\ln(1-x)=- \sum_{k=1}^ \infty \frac{ x^k}{ k} \ \ \ \ \text{ with radius of convergence = 1 }$$
$$L:= \lim_{n \to \infty }\prod_{r=1}^n \frac{n^2-r}{n^2+r}$$
$$\ln(L) =\lim_{n \to \infty } \sum_{r=1}^n \ln\left(1 – \frac{r}{n^2}\right) – \ln\left(1 + \frac{r}{n^2}\right) $$
$$=-2\lim_{n \to \infty } \sum_{r=1}^n \sum_{k\in 2\mathbb{ N}-1} \frac{ \left(\frac{r}{n^2} \right)^k}{ k} = -1$$
$$L=e^{-1}$$
But this proof missing a few details and tried to complete it
$$\lim_{n \to \infty }\frac{\sum_{r=1}^n r^k}{n^{k+1}}=\lim_{n\to \infty }\frac{\sum\limits_{r=1}^n\left(\frac rn\right)^k}n=\int_0^1x^kdx=\frac1{k+1}$$
hence for all $k >1 $ $$\lim_{n \to \infty }\frac{\sum_{r=1}^n r^k}{n^{2k}} =\lim_{n \to \infty }\frac{n^{k+1}}{(k+1)n^{2k}}=\lim_{n \to \infty }\frac{ n ^{1-k} }{k+1}=0.$$
$$\lim_{n \to \infty } \sum_{r=1}^n \sum_{k\in 2\mathbb{N}-1} \frac{ \left(\frac{r}{n^2} \right)^k}{ k}=\lim_{n \to \infty } \sum_{k\in 2\mathbb{N}-1} \sum_{r=1}^n \frac{ \left(\frac{r}{n^2} \right)^k}{ k}=\lim_{n \to \infty } \sum_{k\in 2\mathbb{N}-1} \frac{ \sum\limits_{r=1}^nr^k}{kn^{2k}}$$ $$=\frac{1}{2}+\lim_{n \to \infty } \sum_{k\in 2\mathbb{N}+1}\frac{ \sum\limits_{r=1}^nr^k}{kn^{2k}},$$
and
$$\lim_{n \to \infty } \sum_{k\in 2\mathbb{N}+1}\frac{ \sum\limits_{r=1}^nr^k}{kn^{2k}}\le\lim_{n \to \infty } \sum_{k=3}^\infty\frac{ \sum\limits_{r=1}^nr^k}{n^{2k}}=\sum_{k=3}^\infty\lim_{n \to \infty } \frac{ \sum\limits_{r=1}^nr^k}{n^{2k}}=\sum_{k=3}^\infty0=0.$$
The problem is I am not sure what is the justification of changing the order of limit and sum in the last step. Is my proof complete? and if not, how to complete it?
Best Answer
By noting that $ \log(1-x) - \log(1+x) = -2 \int_{0}^{x} \frac{1}{1-t^2} \, \mathrm{d}t $, you can prove that
$$ -2x \geq \log(1-x) - \log(1+x) \geq -\frac{2x}{1-x^2}. $$
Then for $n \geq 1$ and $0 \leq r \leq n$,
$$ -\frac{2r}{n^2} \geq \log(1-r/n^2) - \log(1+r/n^2) \geq -\frac{2r}{n^2-(r/n)^2} \geq - \frac{2r}{n^2-1}, $$
hence by summing this for $ r = 1, 2, \ldots, n$ we get:
$$ - \frac{n(n-1)}{n^2} \geq \log \prod_{r=1}^{n} \frac{n^2 - r}{n^2 + r} \geq - \frac{n(n-1)}{n^2 - 1}. $$
This is enough to conclude that $\log L = -1$ and hence $L = e^{-1}$.