Limit of a sequence including infinite product. $limlimits_{n toinfty}prod_{k=1}^n left(1+frac{k}{n^2}right)$

I need to find the limit of the following sequence:
$$\lim\limits_{n \to\infty}\prod_{k=1}^n \left(1+\frac{k}{n^2}\right)$$

Answer

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!