Limit of a product with growing number of factors I'm trying to solve the following limit:
$$L=\lim_{n\to\infty}\frac{(n^2-1)(n^2-2)\cdots(n^2-n)}{(n^2+1)(n^2+3)\cdots(n^2+2n-1)}=\lim_{n\to\infty}\prod_{k=1}^n\frac{n^2-k}{n^2+2k-1}$$
Since originally, the limit yields an $1^\infty$ indeterminate expression, my first idea was taking logarithms:
$$\log L = \lim_{n\to\infty}\log\left(\prod_{k=1}^n\frac{n^2-k}{n^2+2k-1}\right)=\\
=\lim_{n\to\infty}\sum_{k=1}^{n}\log\left(\frac{n^2-k}{n^2+2k-1}\right)=\\=\lim_{n\to\infty}\sum_{k=1}^n\log\left(\frac{1-\frac{k}{n^2}}{1+\frac{2k-1}{n^2}}\right)=\\
=\lim_{n\to\infty}\sum_{k=1}^n\left(\log\left(1-\frac{k}{n^2}\right)-\log\left(1+\frac{2k-1}{n^2}\right)\right)=\\=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log\left(1-\frac{k}{n^2}\right)^{n^2}-\log\left(1+\frac{2k-1}{n^2}\right)^{n^2}\right)=\\
=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}(-k)}-\log\left(1+\frac{1}{\frac{n^2}{2k-1}}\right)^{\frac{n^2}{2k-1}(2k-1)}\right)$$
Now if I could do the limit of the terms inside the sum I'd have:
$$\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}(-k)}-\log\left(1+\frac{1}{\frac{n^2}{2k-1}}\right)^{\frac{n^2}{2k-1}(2k-1)}\right)=\\=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log e^{-k}-\log e^{2k-1}\right)=\\
=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(-k-2k+1\right)=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(-3k+1\right)$$
and then the limit would be trivial. The question is: is that allowed? Why? If not, how could I proceed?
Thanks!
 A: Let's take the first part
$$\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}(-k)} = -\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n k\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}}$$
as example. Note that $-\frac{n^2}{k}$ tends to $\infty$ for any $k=1,2,\cdots,n$, so we can use Taylor's expansion
$$\left(1+\frac{1}{x}\right)^x = e - \frac{e}{2x} + O\left(\frac{1}{x^2}\right) \quad (x \to \infty)$$
to obtain
$$\begin{aligned}
\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}(-k)} 
&= -\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n k\log\left[e+\frac{ek}{2n^2}+O\left(\frac{k^2}{n^4}\right)\right] \\
&= -\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n \left[k + k\log\left(1+\frac{k}{2n^2}+O\left(\frac{k^2}{n^4}\right)\right)\right] \\
&= -\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n \left[k + k\left(\frac{k}{2n^2}+O\left(\frac{k^2}{n^4}\right)\right)\right] \\
&= -\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n k\left[1+O\left(\frac{1}{n}\right)\right],
\end{aligned}$$
where we have used $\frac{k}{n^2}=O\left(\frac{1}{n}\right)$. So the term $O(1/n)$ can be discarded.
Note: An alternative solution is given here:
Lemma. Given $f(0) = 0$ and that finite $f'(0)$ exists, let
$$a_n = \sum_{k=1}^n f\left(\frac{k}{n^2}\right) = f\left(\frac{1}{n^2}\right) + f\left(\frac{2}{n^2}\right) + \cdots + f\left(\frac{n}{n^2}\right),$$
then
$$\lim_{n\to\infty} a_n = \frac{f'(0)}{2}.$$
Proof. First notice that
$$f'(0) = \lim_{x \to 0} \frac{f(x)-f(0)}{x-0} = \lim_{x \to 0} \frac{f(x)}{x},$$
and using the definition of limit, $\forall \varepsilon > 0$, $\exists \delta > 0$, s.t. $\forall 0 < x < \delta$,
$$f'(0) - \varepsilon < \frac{f(x)}{x} < f'(0) + \varepsilon.$$
Particularly for $x > 0$, we have $(f'(0)-\varepsilon)x < f(x) < (f'(0)+\varepsilon)x$.
Now pick $N \in \mathbb{N}$, s.t. $N > \dfrac{1}{\delta}$. For $n > N$,
$$\frac{k}{n^2} \leq \frac{1}{n} < \frac{1}{N} < \delta, \quad k=1,2,\cdots,n,$$
and hence
$$(f'(0) - \varepsilon) \cdot \frac{k}{n^2} < f\left(\frac{k}{n^2}\right) < (f'(0) + \varepsilon) \cdot \frac{k}{n^2}, \quad k=1,2,\cdots,n.$$
Taking summation over $k$ gives
$$\frac{f'(0)-\varepsilon}{2} \cdot \frac{n+1}{n} < \sum_{k=1}^n f\left(\frac{k}{n^2}\right) < \frac{f'(0)+\varepsilon}{2} \cdot \frac{n+1}{n}.$$
Since $\dfrac{n+1}{n} \to 1$ as $n \to \infty$, then $\exists N_1 \in \mathbb{N}$, s.t. $\forall n > N_1$,
$$\frac{f'(0)-\varepsilon}{2} - \frac{\varepsilon}{2} < \sum_{k=1}^n f\left(\frac{k}{n^2}\right) < \frac{f'(0)+\varepsilon}{2} + \frac{\varepsilon}{2},$$
that is, $\dfrac{1}{2}f'(0)-\varepsilon < \sum\limits_{k=1}^n f\left(\frac{k}{n^2}\right) < \dfrac{1}{2}f'(0)+\varepsilon$. Therefore we have
$$\lim_{n\to\infty} a_n = \lim_{n\to\infty} \sum\limits_{k=1}^n f\left(\frac{k}{n^2}\right) = \frac{f'(0)}{2}.$$

Return to your problem, we have
$$I = \lim_{n\to\infty}\prod_{k=1}^n\frac{n^2-k}{n^2+2k-1} = \lim_{n\to\infty}\prod_{k=1}^n\frac{n^2-k}{n^2+2k} = \lim_{n\to\infty}\prod_{k=1}^n \frac{1-\frac{k}{n^2}}{1+2\frac{k}{n^2}} = \lim_{n\to\infty} e^{\sum_{k=1}^n f(k/n^2)},$$
where $f(x) = \ln\left(\frac{1-x}{1+2x}\right)$ and the second equality comes from
$$1 \leftarrow \left(1-\frac{1}{n^2}\right)^n \leq \prod_{k=1}^n \frac{n^2+2k-1}{n^2+2k} = \prod_{k=1}^n \left(1-\frac{1}{n^2+2k}\right) \leq 1.$$
Therefore $I = e^{\frac{f'(0)}{2}} = \boxed{e^{-\frac{3}{2}}}$.
A: $\require{cancel}$
$$L=\lim_{n\to\infty}\frac{\cancel{n^{2n}}(1-\frac1{n^2})(1-\frac2{n^2})\cdots(1-\frac{n}{n^2})}{\cancel{n^{2n}}(1+\frac1{n^2})(1+\frac3{n^2})\cdots(1+\frac2n-\frac1{n^2})}=\lim_{n\to\infty}\frac{e^{\frac{-1}{n^2}}e^{\frac{-2}{n^2}}...e^{\frac{-n}{n^2}}}{e^{\frac{1}{n^2}}e^{\frac{3}{n^2}}...e^{\frac{2n-1}{n^2}}}=\lim_{n\to\infty}\frac{e^{\frac{-1}{n^2}{(\sum_{r=1}^n r)}}}{e^{\frac{1}{n^2}{((2\sum_{r=1}^n r)-n)}}} \\=\lim_{n\to\infty}e^{\frac{-1}{n^2}((3\sum_{r=1}^n r)-n)}=\lim_{n\to\infty}e^{((-3\frac{1}{n}\sum_{r=1}^n \frac{r}{n})+\frac1n)}=\lim_{n\to\infty}e^{((-3\int_0^1xdx+\frac1n)}\to e^{\frac{-3}{2}}$$
