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
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Hint: The $n$-th term in the series is $\dfrac{1}{\sum_{k=1}^n k^2}=\dfrac{6}{n(n+1)(2n+1)}$. Use partial fractions.
lhf
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Along with the hint by lhj, do you know Euler's constant? – DanielWainfleet Feb 15 '16 at 23:56
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Wow. I didn't think it would be that easy. – Saikat Feb 17 '16 at 14:13
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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:
$$ \begin{align} \log 2 &= \sum_{n=1}^\infty\left(\frac{1}{2n-1}- \frac{1}{2n}\right) \\ \sum_{n=1}^N \frac{1}{\sum_{r=1}^n r^2} &= \sum_{n=1}^N \frac{6}{n(n+1)(2n+1)} \\ &= 12 \sum_{n=1}^N \left(\frac{1}{2n} + \frac{1}{2n+2} - \frac{2}{2n+1}\right) \\ &= 12 \sum_{n=1}^N \frac{1}{2n} + 12\left(-\frac{1}{2}+\frac{1}{2N+2}+\sum_{n=1}^N \frac{1}{2n}\right) - 24\left(-1+\frac{1}{2N+1}+\sum_{n=1}^N\frac{1}{2n-1}\right) \\ &= 12\left(-\frac{1}{2}+\frac{2}{2N+2}+2+\frac{1}{2N+1}\right) -24\sum_{n=1}^N\left(\frac{1}{2n-1}-\frac{1}{2n}\right) \\ \lim_{N \to \infty}\sum_{n=1}^N \frac{1}{\sum_{r=1}^n r^2} &= 18-24\log 2 \end{align} $$
GEdgar
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1Thanks for your comment and explanation, and for a more rigorous proof. (+1) – Hypergeometricx Feb 17 '16 at 16:16