TITLE: Convergence of $\left( \frac{1}{1} \right)^2+\left( \frac{1}{2}+\frac{1}{3} \right)^2+\cdots$
QUESTION [11 upvotes]: Does the following series converge? If yes, what is its value in simplest form?
$$\left( \frac{1}{1} \right)^2+\left( \frac{1}{2}+\frac{1}{3} \right)^2+\left( \frac{1}{4}+\frac{1}{5}+\frac{1}{6} \right)^2+\left( \frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10} \right)^2+\dots$$
I have no idea how to start. Any hint would be really appreciated. THANKS!

REPLY [4 votes]: Approximation with Euler-Maclaurin Sum Formula
The Euler-Maclaurin Sum Formula gives the fairly well-known asymptotic series for the Harmonic Numbers
$$
\begin{align}
H_n
&=\gamma+\log(n)+\frac1{2n}-\frac1{12n^2}+\frac1{120n^4}\\[3pt]
&-\frac1{252n^6}+\frac1{240n^8}-\frac1{132n^{10}}+O\!\left(\frac1{n^{12}}\right)\tag1
\end{align}
$$
where $\gamma$ is the Euler-Mascheroni constant. $\gamma$ doesn't actually come from the Euler-Maclaurin Sum Formula, but is defined as $\lim\limits_{n\to\infty}(H_n-\log(n))$.
Substituting into $(1)$ and expanding Taylor series gives
$$
\begin{align}
\left(H_{n(n+1)/2}-H_{n(n-1)/2}\right)^2
&=\frac4{n^2}-\frac{16}{3n^4}+\frac{32}{45n^6}+\frac{1424}{315n^8}\\[3pt]
&+\frac{3392}{1575n^{10}}-\frac{112912}{10395n^{12}}+O\!\left(\frac1{n^{14}}\right)\tag2
\end{align}
$$
Equation $(2)$ not only shows that the series converges, but applying the Euler-Maclaurin Sum Formula to $(2)$ results in
$$
\begin{align}
&\sum_{k=1}^n\left(H_{k(k+1)/2}-H_{k(k-1)/2}\right)^2\\
&=C-\frac4n+\frac2{n^2}+\frac{10}{9n^3}-\frac8{3n^4}+\frac{398}{225n^5}+\frac{16}{45n^6}-\frac{4378}{2205n^7}\\[3pt]
&+\frac{712}{315n^8}-\frac{22718}{14175n^9}+\frac{1696}{1575n^{10}}+\frac{138}{4235n^{11}}-\frac{56456}{10395n^{12}}+O\!\left(\frac1{n^{13}}\right)\tag3
\end{align}
$$
Plugging $n=200$ into $(3)$ gives
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^\infty\left(H_{k(k+1)/2}-H_{k(k-1)/2}\right)^2=3.170466061427153684757796531}\tag4
$$
As a check, $n=300$ gives the same result.