Find the value of $\lim_{n\to \infty}\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)^{1/2}\ldots(2)^{1/n}$ Find the value of
$$\lim_{n\to \infty}\bigg(1+\dfrac{1}{n}\bigg)\bigg(1+\dfrac{2}{n}\bigg)^{\frac12}\ldots(2)^{\frac{1}{n}}$$  
My work:
$\bigg(1+\dfrac{1}{n}\bigg)=\bigg\{\bigg(1+\dfrac{1}{n}\bigg)^n\bigg\}^{\frac{1}{n}}=e^{\frac{1}{n}}$
$\bigg(1+\dfrac{2}{n}\bigg)^{\frac12}=\bigg\{\bigg(1+\dfrac{2}{n}\bigg)^{\frac{n}{2}}\bigg\}^{\frac{1}{n}}=e^{2\cdot\frac12\cdot\frac{1}{n}}=e^\frac{1}{n}$
$~~~~~~~~~~~~\vdots$
$~~~~~~~~~~~~\vdots$
$\bigg(1+\dfrac{n}{n}\bigg)^{\frac{1}{n}}=e^{\frac{1}{n}}$
So, $L=e$
But, the answer says $L=e^{\frac{\pi^2}{12}}$.
I do not know where I am going wrong, is the answer a typo or I am doing wrong. Please help.
 A: Let $x$ be the value of the limit .Take $\log(x)$ to get $$\lim_{n\to \infty}\sum_{i=1}^n \frac{1}{i/n}\log(1+i/n)\cdot\frac{1}{n}$$ which is equal to $\int_0^1 \frac{\log(1+x)}{x}\mathrm{d}x$ which is $\pi ^2/12.$
A: This seems to be the reasoning in your argument
$$
\begin{align}
\lim_{n\to\infty}\prod_{k=1}^n\left(1+\frac kn\right)^{1/k}
&=\lim_{n\to\infty}\left(\prod_{k=1}^n\left(1+\frac kn\right)^{n/k}\right)^{1/n}\tag{1}\\
&=\lim_{n\to\infty}\left(\prod_{k=1}^n\lim_{n\to\infty}\left[\left(1+\frac kn\right)^{n/k}\right]\right)^{1/n}\tag{2}\\
&=\lim_{n\to\infty}\left(\prod_{k=1}^n\ e\right)^{1/n}\tag{3}\\[12pt]
&=\ e\tag{4}
\end{align}
$$
All of the steps are fine except $(2)$. It is not, in general, allowed to take the limit of an inner part like that. For example consider
$$
\begin{align}
\lim_{n\to\infty}\left(\frac1n\cdot n\right)
&=\lim_{n\to\infty}\left(\lim_{n\to\infty}\left[\frac1n\right] n\right)\tag{5}\\
&=\lim_{n\to\infty}\left(0\cdot n\right)\tag{6}\\[3pt]
&=\lim_{n\to\infty}\ 0\tag{7}\\[2pt]
&=\ 0\tag{8}
\end{align}
$$
Step $(5)$ is the same as step $(2)$, but that step allows us to show that $1=0$.
To see why this affects your limit adversely, notice that no matter how big $n$ gets in the limit, when $k$ is near $n$, $\left(1+\frac kn\right)^{n/k}$ is close to $2$, not $e$. Thus, the terms of the product are between $2$ and $e$. Not all of them tend to $e$.

What we need to do is use the continuity of $\log(x)$ as viplov_jain suggests.
$$
\begin{align}
\log\left(\lim_{n\to\infty}\prod_{k=1}^n\left(1+\frac kn\right)^{1/k}\right)
&=\lim_{n\to\infty}\log\left(\prod_{k=1}^n\left(1+\frac kn\right)^{1/k}\right)\tag{9}\\
&=\lim_{n\to\infty}\sum_{k=1}^n\frac1k\log\left(1+\frac kn\right)\tag{10}\\
&=\lim_{n\to\infty}\sum_{k=1}^n\frac nk\log\left(1+\frac kn\right)\frac1n\tag{11}\\
&=\int_0^1\frac1x\log(1+x)\,\mathrm{d}x\tag{12}\\
&=\int_0^1\sum_{k=0}^\infty(-1)^k\frac{x^k}{k+1}\,\mathrm{d}x\tag{13}\\
&=\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)^2}\tag{14}\\
&=\frac{\pi^2}{12}\tag{15}
\end{align}
$$
Step $(12)$ uses the idea of approximating a Riemann Sum by an integral. $(15)$ tells us that
$$
\lim_{n\to\infty}\prod_{k=1}^n\left(1+\frac kn\right)^{1/k}=e^{\pi^2/12}\tag{16}
$$
Notice that
$$
2\lt2.27610815162573\doteq e^{\pi^2/12}\lt e\tag{17}
$$
