find $\lim_{n \to \infty} \frac{(n^2+1)(n^2+2) \cdots (n^2+n)}{(n^2-1)(n^2-2) \cdots (n^2-n)}$ 
Find $$\lim_{n \to \infty} \frac{(n^2+1)(n^2+2) \cdots (n^2+n)}{(n^2-1)(n^2-2) \cdots (n^2-n)}$$

I tried to apply the squeeze theorem, yet none of my attempts led me to the solution.
 A: Let $f(n)$ be defined by 
\begin{align*}
f(n) &= \frac{(n^2+1)(n^2+2)\cdots(n^2+n)}{(n^2-1)(n^2-2)\cdots(n^2-n)}\\
&= \frac{(1 + {1 \over n^2} )(1 +{2 \over n^2})\cdots(1 + {n \over n^2})}{(1 - {1 \over n^2} )(1 - {2 \over n^2})\cdots(1 - {n \over n^2})}
\end{align*}
Then 
$$\ln f(n) = \sum_{k=1}^n \ln\left(1 + {k \over n^2}\right) - \sum_{k=1}^n \ln\left(1 - {k \over n^2}\right)$$
Using the Taylor series for $\ln(1 + x)$, this can be rewritten as
$$ \ln f(n) = \sum_{k=1}^n \left({k \over n^2} + O\left({k^2 \over n^4}\right)\right) - \left[\sum_{k=1}^n \left(-{k \over n^2} + O\left({k^2 \over n^4}\right)\right)\right]$$
Using the formulas for $\sum_{k=1}^n k$ and $\sum_{k=1}^n k^2$, the above says that
$$ \ln f(n) =  {n(n + 1) \over n^2} + O\left({1 \over n}\right)$$
Thus $$\lim_{n \to \infty} \ln f(n) = 1$$ We conclude that
$$\lim_{n \to \infty}  f(n) = e^1 = e$$
A: The squeeze theorem fails because the limit is not $1$ which in turn happens because the number of factors is unbounded when $n\to\infty$. This suggests that more precise asymptotics might be needed. To get these, consider that the logarithm of the $n$th ratio is
$$
S_n=\sum_{k=1}^n\log\left(1+\frac{k}{n^2}\right)-\log\left(1-\frac{k}{n^2}\right).
$$
As everybody knows, when $x\to0$, $\log(1+x)=x-\frac12x^2+o(x^2)$. Choose $n$ large enough such that, for every $|x|\leqslant\frac1n$,
$$
x-x^2\leqslant\log(1+x)\leqslant x.
$$
(It happens that $n=2$ suffices, but this is irrelevant.) Then, applying this double inequality to each $\log\left(1\pm\frac{k}{n^2}\right)$ in $S_n$, one gets
$$
\sum_{k=1}^n2\frac{k}{n^2}-\frac{k^2}{n^4}\leqslant S_n\leqslant\sum_{k=1}^n2\frac{k}{n^2}+\frac{k^2}{n^4},
$$
that is, using the exact value $\sum\limits_{k=1}^nk=\frac12n(n+1)$ and the crude estimate $\sum\limits_{k=1}^nk^2\leqslant n^3$,
$$
\frac{n(n+1)}{n^2}-\frac1{n^4}n^3\leqslant S_n\leqslant\frac{n(n+1)}{n^2}+\frac1{n^4}n^3.
$$
In other words,
$$
1\leqslant S_n\leqslant1+\frac2n,
$$
from which the limit of $\mathrm e^{S_n}$ should be easy to write down.
A: Hint: try logarithms (after cancelling by n-times $n^2$). Write the double-sum of mercator series and check what change-of-order of summation gives...
