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

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  ?
