Compute $\lim\limits_{n\to\infty} \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)^{{1}/{2}}\cdots\left(1+\frac{n}{n}\right)^{{1}/{n}}$ Compute the limit:
$$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)^{\frac{1}{2}}\cdots\left(1+\frac{n}{n}\right)^{\frac{1}{n}}$$
 A: Note at the onset that $1+\frac{k}n\leqslant\mathrm e^{k/n}$ for every $k$ hence the $n$th product $P_n$ is such that $P_n\leqslant\mathrm e$, in particular, the sequence $(P_n)_{n\geqslant1}$ is bounded. 
To show that $(P_n)_{n\geqslant1}$ actually converges and to identify its limit, note that, for every $n$, 
$$
\log(P_n)=\frac1n\sum\limits_{k=1}^nf\left(\frac{k}n\right),
\qquad\text{with}\quad f(x)=\frac{\log(1+x)}x.
$$
The function $f$ is continuous on $[0,1]$ (define $f(0)=1$) hence its Riemann sums converge to its integral and $P_n\to\mathrm e^\ell$ with
$$
\ell=\int_0^1f(x)\mathrm dx=\int_0^1\left(\sum_{n\geqslant1}(-1)^{n+1}\frac{x^{n-1}}n\right)\mathrm dx=\sum_{n\geqslant1}\frac{(-1)^{n+1}}{n^2}=\frac{\pi^2}{12}.
$$
A: We want to calculate $$\lim_{n \to \infty} \prod_{1 \leqslant k \leqslant n} \left(1 + \frac {k} {n}\right)^{\frac {1} {k}}.$$ Denote it by $\ell$. Taking logarithms we have
$$\begin {eqnarray}
\log \ell & = & \lim_{n \to \infty} \sum_{1 \leqslant k \leqslant n} \frac {1} {k} \log \left(1 + \frac {k} {n}\right) \nonumber \\
& = & \lim_{n \to \infty} \sum_{1 \leqslant k \leqslant n} \frac {1} {k} \sum_{m \geqslant 1} (-1)^{m + 1} \frac {k^m} {m n^m} \nonumber \\
& = & \lim_{n \to \infty} \sum_{1 \leqslant k \leqslant n} \sum_{m \geqslant 1} (-1)^{m + 1} \frac {k^{m - 1}} {m n^m} \nonumber \\
& = & \lim_{n \to \infty} \sum_{m \geqslant 1} \sum_{1 \leqslant k \leqslant n} (-1)^{m + 1} \frac {k^{m - 1}} {m n^m} \nonumber \\
& = & \lim_{n \to \infty} \sum_{m \geqslant 1} \left(\frac {(-1)^{m + 1}} {m^2} + O \left(\frac {1} {n}\right) \right) \nonumber \\
& = & \sum_{m \geqslant 1} \frac {(-1)^{m + 1}} {m^2} \nonumber \\
& = & \frac {\pi^2} {12}.
\end {eqnarray}$$
Hence, $\ell = \exp \left(\frac {\pi^2} {12}\right)$.
Note that in step 2 we changed $\log \left(1 + \frac {k} {n}\right)$ with its Taylor expansion, and in step 5 we used the fact that $\sum_{1 \leqslant k \leqslant n} k^p = \frac {n^{p + 1}} {p + 1} + O (n^p)$ for any non-negative integer $p$.
