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.

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
$$\lim_{n \to \infty}  f(n) = e^1 = e$$
