$\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+...\to ?\;\;$ (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 .
 A: 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 $\ln(x+1)$.
A: 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)
$$S=6 \left(2 \gamma -1+2 \psi \left(\frac{3}{2}\right)\right)=6 (3-2 \log 4)$$
A: 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  ?
