<s>[INST] The sum of an infinite series containing a finite series in each denominator 
Evaluate
  $$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{10^2}+\cdots+\frac{1}{\left[\frac{k(k+1)}{2}\right]^2}+\cdots$$
   to $\infty$, where $k$ is the $k$th term of the series.

Using Microsoft Excel, I found that the sum of the first $100$ terms $=1.15947\dots$.
Does the convergence value have an exact form? If yes, what is it?
 [/INST] $$\frac {1}{k(k+1)}= \frac {1}{k}-\frac {1}{k+1}$$
$$\frac {1}{k^2(k+1)^2}= \frac {1}{k^2}+\frac {1}{(k+1)^2} - \frac {2}{k(k+1)}$$
$$\sum_1^{\infty}\frac {1}{k^2(k+1)^2}= \sum_1^{\infty}\frac {1}{k^2}+\sum_1^{\infty}\frac {1}{(k+1)^2} -\sum_1^{\infty} \frac {2}{k(k+1)}=$$
$$\frac {\pi ^2}{6} +(\frac {\pi ^2}{6}-1)-2$$
$$\sum_1^{\infty} \frac {4}{k^2(k+1)^2} = 4(\frac {\pi^2 }{6} + \frac {\pi^2 }{6}-1 -2) = 4(\frac {\pi^2-9}{3})\approx 1.159472535...$$
</s>