<s>[INST] 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? 
 [/INST] Let $\mathcal{R}_n$ denote the integral on the right-hand-side of eq. (2):
$$
    \mathcal{R}_n = \int_0^1 2 x^{n-1} \log\left(1+\frac{1}{\sqrt{x}}\right) \mathrm{d}x
$$
Consider
$$
\begin{eqnarray}
 (n+1) \mathcal{R}_{n+1} - n \mathcal{R}_n &=& \int_0^1 2 \left( (n+1) x^{n} - n x^{n-1} \right) \log\left(1+\frac{1}{\sqrt{x}}\right) \mathrm{d}x \\
   &=& \int_0^1 2 \log\left(1+\frac{1}{\sqrt{x}}\right) \mathrm{d} \left( -x^n \left(1-x\right)\right) \\
    &=&  \int_0^1 \left(\sqrt{x}-1\right)x^{n-1} \mathrm{d} x = \frac{2}{2n+1} - \frac{1}{n} 
\end{eqnarray}
$$
Therefore:
$$
   n \mathcal{R}_n = \mathcal{R}_1 + \sum_{m=1}^{n-1} \left( \frac{2}{2m+1} - \frac{1}{m} \right) = \psi\left(n+\frac{1}{2}\right) - \psi(n) + 2 \left(\log(2)-1\right) + \mathcal{R}_1
$$
Integral $\mathcal{R}_1$ can be easily integrated by parts:
$$
  \mathcal{R}_1 = 2 \int_0^1 \log\left(1+\frac{1}{\sqrt{x}}\right)\mathrm{d} x\stackrel{\text{by parts}}{=} \left. 2\left(  x \log\left( 1 + \frac{1}{\sqrt{x}}\right) + \sqrt{x} - \log\left(1+\sqrt{x}\right) \right) \right|_{x \downarrow 0}^{x = 1} = 2
$$
Thus
$$
  n \mathcal{R}_n = 2 \log(2) + \psi\left(n+\frac{1}{2}\right) - \psi(n)
$$
Similarly, denoting $\mathcal{L}_n = \int_0^1 f_n(x)^2 \mathrm{d} x$ the integral on the left-hand-side of eq. (2):
$$ \begin{eqnarray}
  n \left(\mathcal{L}_{n+1} - \mathcal{L}_n\right) &=& \int_0^1 \left( n f_{n+1}(x)^2 - n \left( \frac{x^n}{n} + f_{n+1}(x) \right)^2 \right)\mathrm{d} x \\
   &=& \int_0^1 \left( -2 x^n f_{n+1}(x) - \frac{x^{2n}}{n}\right)\mathrm{d} x \\
   &=& \color\maroon{2 \int_0^1 x^n \log(1-x) \mathrm{d} x} + {\color\blue{2 \int_0^1 x^n \sum_{k=1}^{n} \frac{x^k}{k} \mathrm{d} x}} - \frac{1}{n(2n+1)} \\
   &=& \color\maroon{-\frac{2}{n+1} H_{n+1}} + 2 \sum_{k=1}^n \frac{1}{k(k+n+1)} - \frac{1}{n(2n+1)} \\
   &=& \frac{2}{n+1} \left( \psi(n+1) - \psi(2+2n) \right) - \frac{1}{n(2n+1)} \\
   &=& \frac{1}{n (n+1)} \left( \psi(n+1)-\psi\left(n+\frac{3}{2}\right)-2 \log (2) \right) -\frac{1}{n^2 (2 n+1)}
\end{eqnarray}
$$
where $f_n(x) = \sum_{k=n}^\infty \frac{x^k}{k}$. Now since $\mathcal{L}_1 = \int_0^1 \log^2(1-x)\mathrm{d} x = 2$, and since $\mathcal{R}_{n+1} - \mathcal{R}_n$ equals to 
$\mathcal{L}_{n+1} -\mathcal{L}_n$, we establish that $\mathcal{L}_n = \mathcal{R}_n$:
$$
  \mathcal{R}_{n+1} - \mathcal{R}_n = 2\log(2) \left(  \frac{1}{n+1}- \frac{1}{n} \right) 
   + \frac{1}{n+1} \left( \psi\left(n+\frac{3}{2}\right) - \psi (n+1)\right) - 
     \frac{1}{n} \left( \psi\left(n+\frac{1}{2}\right) - \psi (n)\right) \stackrel{\color\maroon{\text{use recurrence equation for } \psi}}{=} \mathcal{L}_{n+1} - \mathcal{L}_n
$$
</s>