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!

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.
