<s>[INST] $\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+...\to ?\;\;$ (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 .
 [/INST] 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)
$$S=6 \left(2 \gamma -1+2 \psi \left(\frac{3}{2}\right)\right)=6 (3-2 \log 4)$$
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