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

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 $\ln(x+1)$.

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) $$S=6 \left(2 \gamma -1+2 \psi \left(\frac{3}{2}\right)\right)=6 (3-2 \log 4)$$