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?
 A: $$\frac {1}{k(k+1)}= \frac {1}{k}-\frac {1}{k+1}$$
$$\frac {1}{k^2(k+1)^2}= \frac {1}{k^2}+\frac {1}{(k+1)^2} - \frac {2}{k(k+1)}$$
$$\sum_1^{\infty}\frac {1}{k^2(k+1)^2}= \sum_1^{\infty}\frac {1}{k^2}+\sum_1^{\infty}\frac {1}{(k+1)^2} -\sum_1^{\infty} \frac {2}{k(k+1)}=$$
$$\frac {\pi ^2}{6} +(\frac {\pi ^2}{6}-1)-2$$
$$\sum_1^{\infty} \frac {4}{k^2(k+1)^2} = 4(\frac {\pi^2 }{6} + \frac {\pi^2 }{6}-1 -2) = 4(\frac {\pi^2-9}{3})\approx 1.159472535...$$
A: Just written for your curiosity.
As showed in answers, the infinit sum is not to hard. We could also have a good approximation  of the partial sum
$$S_p=\sum_{k=1}^p\frac{4}{k^2 (k+1)^2}$$ Using a CAS,
$$S_p=\frac{4 \left((p+1)^2 \left(\pi ^2-6 \psi ^{(1)}(p+1)\right)-3 p (3 p+4)\right)}{3
   (p+1)^2}$$ and using the asymptotics of the polygamma function
$$S_p=\left(\frac{4 \pi ^2}{3}-12\right)-\frac{4}{3
   p^3}+\frac{4}{p^4}+O\left(\frac{1}{p^5}\right)$$
For $p=10$, the exact value is $\frac{22254209}{19209960}\approx 1.15847$ while the above approximation gives $\frac{4 \pi ^2}{3}-\frac{90007}{7500}\approx 1.15854$.
A: 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).
$$
=4\left(-1+ \sum_{k=1}^\infty \frac 1{k^2}
\right) +
4\left( \sum_{k=1}^\infty \frac 1 {k^2}
\right)+
8\left( \sum_{k=1}^\infty  \frac 1{(k + 1)}- \frac 1 k
\right)
\\=
-4+4\cdot \frac{\pi^2} 6 + 4\cdot \frac{\pi^2} 6 -8
=
4·(\pi^2 - 9)/3
$$
