<s>[INST] Is there a closed-form of $\frac{1}{1}+\frac{1}{1+2^2}+\frac{1}{1+2^2+3^2}+.....$ How can I find the closed-form of?
$$\frac{1}{1}+\frac{1}{1+2^2}+\frac{1}{1+2^2+3^2}+.....$$
Any help thanks
 [/INST] hypergeometric has a good idea, but we cannot work with divergent series this way.  Similar rearrangements can give wrong answers.  But taking hypergeometric's ideas, a valid proof looks like this:
$$
\begin{align}
\log 2 &= \sum_{n=1}^\infty\left(\frac{1}{2n-1}- \frac{1}{2n}\right)
\\
\sum_{n=1}^N \frac{1}{\sum_{r=1}^n r^2} &= 
\sum_{n=1}^N \frac{6}{n(n+1)(2n+1)}
\\ &=
12 \sum_{n=1}^N \left(\frac{1}{2n} + \frac{1}{2n+2} - \frac{2}{2n+1}\right)
\\ &=
12 \sum_{n=1}^N \frac{1}{2n} + 12\left(-\frac{1}{2}+\frac{1}{2N+2}+\sum_{n=1}^N \frac{1}{2n}\right)
- 24\left(-1+\frac{1}{2N+1}+\sum_{n=1}^N\frac{1}{2n-1}\right)
\\ &=
12\left(-\frac{1}{2}+\frac{2}{2N+2}+2+\frac{1}{2N+1}\right)
-24\sum_{n=1}^N\left(\frac{1}{2n-1}-\frac{1}{2n}\right)
\\
\lim_{N \to \infty}\sum_{n=1}^N \frac{1}{\sum_{r=1}^n r^2} &=
18-24\log 2
\end{align}
$$
</s>