Convergence of $\left( \frac{1}{1} \right)^2+\left( \frac{1}{2}+\frac{1}{3} \right)^2+\cdots$ 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!
 A: Since for moderately large values of $n$ we have
$$ H_n \approx \log(n)+\gamma+\frac{1}{2n}-\frac{1}{12n^2} $$
we also have
$$ H_{n(n+1)/2}-H_{n(n-1)/2} \approx \frac{2}{n}-\frac{4}{3n^3}$$
and
$$ \sum_{n\geq 1}\left[H_{n(n+1)/2}-H_{n(n-1)/2}\right]^2 \approx 1+\sum_{n\geq 2}\left(\frac{2}{n}-\frac{4}{3n^3}\right)^2=1+\frac{2 \left(-1890+2835 \pi ^2-252 \pi ^4+8 \pi ^6\right)}{8505} $$
is finite and approximately equal to $\color{green}{3.17}151$. I doubt there is a simple closed form.
A: Notice, that:
$$\left(\frac{1}{1}\right)^2+\left(\frac{1}{2} + \frac{1}{3}\right)^2 + \dots<\left(\frac{1}{1}\right)^2+\left(\frac{2}{2}\right)^2+\left(\frac{3}{4}\right)^2+\left(\frac{4}{7}\right)^2=2+\sum_{n=2}^{\infty}\left(\frac{n}{\frac{n(n-1)}{2}+1}\right)^2<2+\sum_{n=2}^{\infty}\left(\frac{2}{n-1}\right)^2$$
Since it's bounded by converging series, it's convergent itself.
A: Suggestion.
This is merely a suggestion that is too large to fit in a comment. Euler's integral formula for the harmonic numbers, $H_n=\int_0^1\frac{1-x^n}{1-x}\ \mathrm{d}x$, gives us a formula for the series as $a\to1^-$:
$$S=4\sum_{n=1}^{\infty}\left[f_a(n(n+1))-f_a(n(n-1))\right]^2$$
where $f_a(x)=\int_{0}^{a}\frac{t^{x+1}}{1-t^{2}}\ \mathrm{d}t$. The function, $f_a$, can also be given in terms of the hypergeometric function, as $\frac{a^{x+2}}{x+2}\cdot{_2F}_1\left(1,\frac{x}2+1;\frac{x}2+2;a^{2}\right)$.
Thus, it could potentially aid in finding an analytic solution to $S$ (or showing whether that would be impossible) by finding a closed form for the square of the difference of hypergeometric functions, $\left[f_a(n(n+1))-f_a(n(n-1))\right]^2$. However, no such solution has leapt out to me so far.
