How to prove that $\int_0^1\left(\sum\limits_{k=n}^\infty {x^k\over k}\right)^2\,dx = \int_0^1 2x^{n-1}\log\left(1+{1\over\sqrt{x}}\right)\,dx$ American Mathematical Monthly problem 11611 essentially asks you to show that 
$$\lim_n\ n  \int_0^1\left(\sum_{k=n}^\infty {x^k\over k}\right)^2\,dx=2\log(2).\tag1$$ 
This would follow easily from (2) below, which is true for small values of $n$
 according to Maple. 
But I couldn't prove equation (2) in general, so I found a direct solution for (1) instead.   
$$\int_0^1\left(\sum_{k=n}^\infty {x^k\over k}\right)^2\,dx = \int_0^1 2x^{n-1}\log\left(1+{1\over\sqrt{x}}\right)\,dx.\tag2$$
But I'm still curious about (2). What am I missing? How can equation (2) be proven? 

Here's a way to do it by brute force. First write
$$
\begin{align}
\int_0^1\left(\sum_{k = n}^\infty \frac{x^k}{k}\right)^2\,dx & = \sum_{k,m\geq n}\frac{1}{km}\int_0^1x^{k+m}\,dx \\
& = \sum_{k,m \geq n} \frac{1}{km(k+m+1)}.
\end{align}
$$
Put $r = k+m$, so that $m = r-k$, and transform the sum:
$$
\begin{align}
\sum_{k,m\geq n} \frac{1}{km(k+m+1)} &= \sum_{r = 2n}^\infty \frac{1}{r+1}\sum_{k = n}^{r-n}\frac{1}{k(r-k)} \\
&= \sum_{r = 2n}^\infty \frac{1}{r(r+1)}\sum_{k=n}^{r-n}\left(\frac{1}{k} + \frac{1}{r-k}\right) \\
& = \sum_{r = 2n}^\infty \frac{2}{r(r+1)}\sum_{k = n}^{r-n}\frac{1}{k} \\
& = 2\sum_{r = 2n}^\infty\left(\frac{1}{r}\sum_{k = n}^{r-n}\frac{1}{k} - \frac{1}{r+1}\sum_{k = n}^{r-n}\frac{1}{k}\right) \\
& = 2\sum_{r = 2n}^\infty\left(\frac{1}{r}\sum_{k = n}^{r - n}\frac{1}{k} - \frac{1}{r+1}\sum_{k = n}^{r+1 - n} \frac{1}{k}\right) + 2 \sum_{r = 2n}^\infty \frac{1}{(r+1)(r+1 - n)}. \\
\end{align}
$$
The first sum in the last line telescopes, so it can be evaluated as
$$
2\sum_{r = 2n}^\infty \left(\frac{1}{r}\sum_{k = n}^{r - n}\frac{1}{k} - \frac{1}{r+1}\sum_{k = n}^{r+1 - n}\frac{1}{k}\right) = \lim_{r \to \infty} \left(\frac{1}{n^2} - \frac{2}{r+1}\sum_{k = n}^{r+1 - n}\frac{1}{k}\right) = \frac{1}{n^2}.
$$
Thus we need to prove that
$$
\frac{1}{n^2} + 2\sum_{r = 2n}^\infty \frac{1}{(r+1)(r+1 - n)} = 2\int_0^1 x^{n-1}\log(1+x^{-1/2})\,dx. $$
Since
$$
\begin{align}
2\int_0^1 x^{n-1}\log(1+x^{-1/2})\,dx &= 2\int_0^1 x^{n-1}\log(1 + x^{1/2})\,dx - \int_0^1 x^{n-1}\log{x}\,dx \\
& = 2\int_0^1 x^{n-1}\log(1+x^{1/2})\,dx + \frac{1}{n^2},
\end{align}
$$
we need only prove that
$$
\sum_{r = 2n}^\infty \frac{1}{(r+1)(r+1 - n)} = \int_0^1 x^{n-1}\log{(1+x^{1/2})}\,dx.
$$
This can be done by developing $\log{(1+x^{1/2})}$ in powers of $x^{1/2}$:
$$
\begin{align}
\int_0^1 x^{n-1}\log{(1+x^{1/2})}\,dx & = \sum_{m = 1}^\infty\frac{(-1)^{m+1}}{m} \int_0^1 x^{n- 1 + m/2}\,dx \\ 
& = 2\sum_{m = 1}^\infty \frac{(-1)^{m+1}}{m(2n+m)} \\
& = 2 \sum_{\text{$m$ odd}} - 2\sum_{\text{$m$ even}} \frac{1}{m(2n+m)} \\
& = 2 \sum - 4\sum_{\text{$m$ even}} \frac{1}{m(2n+m)} \\
& = 2 \sum_{m = 1}^\infty \frac{1}{m(2n+m)} - \sum_{m = 1}^\infty \frac{1}{m(n+m)} \\
& = 2 \sum_{m = 1}^\infty \frac{1}{m}\left(\frac{1}{2n+m} - \frac{1}{2n+2m}\right) \\
& =  \sum_{m = 1}^\infty \frac{1}{(2n+m)(n+m)} \\
& = \sum_{r = 2n}^\infty \frac{1}{(r + 1)( r +1 - n)},
\end{align}
$$
which proves the identity. The other methods are obviously more concise, but after working through this I couldn't resist posting it.
