<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] Using fractional decomposition we get:
$$
\frac 1{\left(\frac{k(k+1)}{2}\right)^2} = \frac 4{(k + 1)^2} + \frac 8{(k + 1)} +\frac 4 {k^2} - \frac 8 k
$$
So, your sum simplifies to one telescoping sum, and the sum
$$\sum_{k=1}^\infty \frac 1 {k^2} =\frac{\pi^2} 6$$
So, all together we get:
$$
\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{10^2}+\dots+\frac{1}{[\frac{k(k+1)}{2}]^2}+\dots\\ =
\sum_{k=1}^\infty \frac 1{\left(\frac{k(k+1)}{2}\right)^2}
\\=
\sum_{k=1}^\infty 
\frac 4{(k + 1)^2} + \frac 8{(k + 1)} +\frac 4 {k^2} - \frac 8 k
\\ = 
4\left( \sum_{k=1}^\infty \frac 1{(k + 1)^2}
\right) +
4\left( \sum_{k=1}^\infty \frac 1 {k^2}
\right)+
8\left( \sum_{k=1}^\infty  \frac 1{(k + 1)}- \frac 1 k
\right)
$$
In the first sum, you do an index shift to the left and add  the first summand (as a zero).
$$
=4\left(-1+ \sum_{k=1}^\infty \frac 1{k^2}
\right) +
4\left( \sum_{k=1}^\infty \frac 1 {k^2}
\right)+
8\left( \sum_{k=1}^\infty  \frac 1{(k + 1)}- \frac 1 k
\right)
\\=
-4+4\cdot \frac{\pi^2} 6 + 4\cdot \frac{\pi^2} 6 -8
=
4·(\pi^2 - 9)/3
$$
</s>