$\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+...\to ?\;\;$ (Click here.)

Question

$$\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+\dots\to~?$$ I tried like below $$\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+\dots=\\\sum_{n=1}^{\infty}\frac{1}{1^2+2^2+3^2+\dots+n^2}=\\\sum_{n=1}^{\infty}\frac{1}{\frac{n(n+1)(2n+1)}{6}}=\\\sum_{n=1}^{\infty}\frac{6}{n(n+1)(2n+1)}=\\ $$ Then I can use the fraction ,but $$\frac{1}{n(n+1)(2n+1)}=\frac{1}{n}+\frac{1}{n+1}+\frac{-4}{2n+1}$$ This is ugly to turn into telescopic series . Can you help me to find :series converge to ?

Thanks in advance .

Answer 1

Trying to avoid Daniel Fischer's approach.

$$\frac2{(2n)(2n+1)(2n+2)}=\frac1{(2n)(2n+1)}-\frac1{(2n+1)(2n+2)}$$

Collect the terms by their signs:

$$S=12\sum_{n=2}^\infty\frac{(-1)^n}{n(n+1)}$$

Now apply partial fractions on this nicely and it becomes

$$S=12\sum_{n=2}^\infty\frac{(-1)^n}n+\frac{(-1)^{n+1}}{n+1}$$

You can then collect the terms again to get

$$S=6+12\sum_{n=3}^\infty\frac{(-1)^n}n$$

Upon which you may use the Maclaurin expansion of $\ln(x+1)$.

Answer 2

When I post it , I find an idea ... $$\sum_{n=1}^{\infty}\frac{6}{n(n+1)(2n+1)}=\\ \sum_{n=1}^{\infty}\frac{6\times 2 \times 2}{(2n)(2n+2)(2n+1)}=\\ \sum_{n=1}^{\infty}\frac{24}{(2n)(2n+2)(2n+1)}=\\ \sum_{n=1}^{\infty}\frac{24}{1}(\frac{1}{(2n)(2n+1)(2n+2)})=\\ \sum_{n=1}^{\infty}\frac{24}{2}(\frac{1}{(2n)(2n+1)}-\frac{1}{(2n+1)(2n+2)})=\\ 12\times \frac{1}{2\times 3}=2$$ but wolfram says $$\sum_{n=1}^{\infty}\frac{6}{n(n+1)(2n+1)}=6(3 - 4log(2))$$ Is there something I had missed ?

Answer 3

Starting from $$S=6 \sum _{n=1}^{\infty } \left(\frac{1}{n+1}-\frac{4}{2 n+1}+\frac{1}{n}\right)$$ I changed the index $n\to n+1$ $$S=6 \sum _{n=0}^{\infty } \left(\frac{1}{n+2}-\frac{2}{n+\frac{3}{2}}+\frac{1}{n+1}\right)$$ Then I applied an amazing property of digamma function that can be found here and I got

$$S=6 \left(-\psi (1)-\psi (2)+2 \psi \left(\frac{3}{2}\right)\right)$$ which gives ($\gamma $ is Euler-Mascheroni constant) $$S=6 \left(2 \gamma -1+2 \psi \left(\frac{3}{2}\right)\right)=6 (3-2 \log 4)$$