Limit of a sequence including infinite product. $\lim\limits_{n \to\infty}\prod_{k=1}^n \left(1+\frac{k}{n^2}\right)$ I need to find the limit of the following sequence:
$$\lim\limits_{n \to\infty}\prod_{k=1}^n \left(1+\frac{k}{n^2}\right)$$
 A: Note that $$\prod_{k\leq n}\left(1+\frac{k}{n^{2}}\right)=\prod_{k\leq n}\left(\frac{n^{2}+k}{n^{2}}\right)=\frac{\left(n^{2}+1\right)_{n}}{n^{2n}}
 $$ where $\left(x\right)_{m}
 $ is the Pochhammer symbol and since $$\left(x\right)_{m}=\frac{\Gamma\left(x+m\right)}{\Gamma\left(x\right)}
 $$ we have $$\prod_{k\leq n}\left(1+\frac{k}{n^{2}}\right)=\frac{\Gamma\left(n^{2}+n+1\right)}{n^{2n}\Gamma\left(n^{2}+1\right)}\longrightarrow\color{red}{\sqrt{e}}
 $$ where the last limit can be calculated using the Stirling's approximation.
A: Hint
$$P_n=\prod_{k=1}^n (1+\frac{k}{n^2})\implies \log(P_n)=\sum_{k=1}^n \log\left(1+\frac{k}{n^2}\right)$$ Work with the sum (without forgetting by the end that $P_n=e^{\log(P_n)}$)
A: $$\left(1+\frac{k}{n^2} \right)\left(1+\frac{n-k}{n^2} \right)=\left(1+\frac{1}{n} +\frac{k(n-k)}{n^4} \right)$$
Then we have
$$\left(1+\frac{1}{n}  \right) \leq \left(1+\frac{1}{n} +\frac{k(n-k)}{n^4} \right) \leq \left(1+\frac{1}{n} +\frac{1}{n^2} \right)=\left(1 +\frac{n+1}{n^2} \right)$$
Therefore
$$\left(1+\frac{1}{n}  \right) ^\frac{n}{2} \leq  \prod_{k=1}^n\left(1+\frac{k}{n^2}\right) \leq \left(1 +\frac{n+1}{n^2} \right)^\frac{n}{2}$$
and squeeze it.
A: 
PRIMER:
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1} \tag 1$$ 
for $x>0$.


Note that we have
$$\begin{align}
\log\left(\prod_{k=1}^n \left(1+\frac{k}{n^2}\right)\right)&=\sum_{k=1}^n \log\left(1+\frac{k}{n^2}\right)\tag 2
\end{align}$$
Applying the right-hand side inequality in $(1)$ to $(2)$ reveals
$$\begin{align}
\sum_{k=1}^n \log\left(1+\frac{k}{n^2}\right)&\le \sum_{k=1}^n \frac{k}{n^2}\\\\
&=\frac{n(n+1)}{2n^2} \\\\
&=\frac12 +\frac{1}{2n}\tag 3
\end{align}$$
Applying the left-hand side inequality in $(1)$ to $(2)$ reveals
$$\begin{align}
\sum_{k=1}^n \log\left(1+\frac{k}{n^2}\right)&\ge \sum_{k=1}^n \frac{k}{k+n^2}\\\\
&\ge \sum_{k=1}^n \frac{k}{n+n^2}\\\\
&=\frac{n(n+1)}{2(n^2+n)} \\\\
&=\frac12 \tag 4
\end{align}$$
Putting $(2)-(4)$ together yields
$$\frac12 \le \log\left(\prod_{k=1}^n \left(1+\frac{k}{n^2}\right)\right)\le \frac12+\frac{1}{2n} \tag 5$$
whereby application of the squeeze theorem to $(5)$ gives
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty} \log\left(\prod_{k=1}^n \left(1+\frac{k}{n^2}\right)\right)=\frac12}$$
Hence, we find that 
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\prod_{k=1}^n \left(1+\frac{k}{n^2}\right)=\sqrt e}$$
And we are done!
