On the convergence of the sequence $\,a_n=\displaystyle\sum_{k=1}^{n-1}\left(\frac{n}{k(n-k)}\right)^{2}$ Let the sequence
$$
a_n=\sum_{k=1}^{n-1}\left(\frac{n}{k(n-k)}\right)^{2}, \quad n\ge 2.
$$
Show that:
A. $\,a_n\le 4$.
B. $\,\{a_n\}_{n\in\mathbb N}$ is decreasing, and hence converging.
C. Find $\lim_{n\to\infty}a_n$.
Clearly, B implies A.
The first thing that comes to one's mind is to make $\,\{a_n\}_{n\in\mathbb N}$ look like an approximation of a definite integral. However, we do not have the right power of $n$. 
Numerical experiments suggest that $\lim_{n\to\infty}=3.2898\ldots$ 
 A: Note that $\frac{n}{k(n-k)} = \frac 1k + \frac1{n-k}$, therefore
$$a_n = \sum_{k=1}^{n-1} \left(\frac 1k + \frac1{n-k} \right)^2 = \sum_{k=1}^{n-1} \left(\frac 1{k^2} + \frac1{(n-k)^2} +\frac{2}{k(n-k)}\right) = 2\sum_{k=1}^{n-1} \frac 1{k^2} + \frac 2n\sum_{i=1}^{n-1} \left(\frac 1k + \frac 1{n-k}\right) = 2\sum_{k=1}^{n-1} \frac 1{k^2} + \frac 2n\sum_{k=1}^{n-1} \frac 1k$$
For the monotonicity of $a_n$ check Jack's answer.
Now observe that the limit of $2\sum_{k=1}^{n-1} \frac 1{k^2}$ as $n$ tends to infinity is $2 \cdot \frac {\pi^2}{6} = \frac{\pi^2}{3}$, and the limit of $\frac 2n\sum_{k=1}^{n-1} \frac 1k$ is zero. The first limit is well known, for the second you can use e.g. Stolz theorem.
A: $$\begin{align} 
a_n= & \sum_{k=1}^{n-1}\left[\frac{n}{k(n-k)}\right]^{2} \\ 
= &  \sum_{k=1}^{n-1}\left[\frac{(n-k)+k}{k(n-k)}\right]^{2} \\
= &  \sum_{k=1}^{n-1}\left[\frac{1}{k}+\frac{1}{n-k}\right]^{2} \\
= &  \sum_{k=1}^{n-1}\left[\frac{1}{k^{2}}+\frac{1}{(n-k)^{2}}+\frac{2}{k(n-k)}\right] \\
= &  \sum_{k=1}^{n-1}\frac{1}{k^{2}}+ \sum_{k=1}^{n-1}\frac{1}{(n-k)^{2}}+2 \sum_{k=1}^{n-1}\frac{1}{k(n-k)} \\
= &  \sum_{k=1}^{n-1}\frac{1}{k^{2}}+ \sum_{k=1}^{n-1}\frac{1}{(n-k)^{2}}+\frac{2}{n} \sum_{k=1}^{n-1}\left[\frac{1}{k}+\frac{1}{n-k}\right] \\
= &  \sum_{k=1}^{n-1}\frac{1}{k^{2}}+ \sum_{k=1}^{n-1}\frac{1}{(n-k)^{2}}+\frac{2}{n} \sum_{k=1}^{n-1}\frac{1}{k}+\frac{2}{n} \sum_{k=1}^{n-1}\frac{1}{n-k} \\
\end{align}$$
Now carefully note that the terms in the sum $\sum_\limits{k=1}^{n-1}\frac{1}{k^{2}}$ are same as those in $\sum_\limits{k=1}^{n-1}\frac{1}{(n-k)^{2}}$ only in the reverse order. Same also holds for $\sum_\limits{k=1}^{n-1}\frac{1}{k}$ and $\sum_\limits{k=1}^{n-1}\frac{1}{n-k}$ .
So we can write that
$$\begin{align} 
a_n= &  \sum_{k=1}^{n-1}\frac{2}{k^{2}}+\frac{2}{n} \sum_{k=1}^{n-1}\frac{2}{k} \\
= & 2 \sum_{k=1}^{n-1}\left[\frac{1}{k^{2}}+\frac{2}{nk}\right] \\
\end{align}$$
And as $n\to\infty$ , $\sum_{k=1}^{n-1}\frac{1}{k^{2}} \to \frac{\pi^2}{6}$ and $\sum_{k=1}^{n-1}\frac{2}{nk} \to 0$.
So the limit of $a_n$ is $\frac{\pi^2}{3}$.
A: $$a_n = \sum_{k=1}^{n-1}\left(\frac{1}{k}+\frac{1}{n-k}\right)^2 = 2 H_{n-1}^{(2)}+\frac{2}{n}\sum_{k=1}^{n-1}\left(\frac{1}{k}+\frac{1}{n-k}\right)=2H_{n-1}^{(2)}+\frac{4H_{n-1}}{n}$$
and:
$$ a_{n+1}-a_n = \frac{2}{n^2}+4\left(\frac{H_n}{n+1}-\frac{H_{n-1}}{n}\right)=\frac{6}{n^2}-\frac{4H_n}{n(n+1)}$$
is negative as soon as $4n H_n > 6n+6$. It follows that $\{a_n\}_{n\geq 2}$ is a decreasing sequence converging to $2\,\zeta(2)=\frac{\pi^2}{3}<\frac{10}{3}$. Here, as usual,
$$ H_n = \sum_{k=1}^{n}\frac{1}{k},\qquad H_{n}^{(2)}=\sum_{k=1}^{n}\frac{1}{k^2} $$
and by the Cauchy-Schwarz inequality $H_n\leq \sqrt{n\,\zeta(2)}$.
