how to solve double summation $1/k^2$ How should I solve
$$\sum^{\infty}_{n=1}\sum^n_{k=1} \frac{1}{k^2} $$
I know that $\sum^\infty_{k=1} \frac{1}{k^2} = \frac{\pi^2}{6}$

This is probably off-topic.
In answers, you have been explained that the infinite summation does not converge and why it does not.
Let us look at what happens $$\sum^n_{k=1} \frac{1}{k^2}=H_n^{(2)}$$ where appear the generalized harmonic numbers. So, using their properties, $$S_p=\sum^{p}_{n=1}\sum^n_{k=1} \frac{1}{k^2}=\sum^{p}_{n=1}H_n^{(2)}=(p+1) H_{p+1}^{(2)}-H_{p+1}$$ Considering their asymptotics, we then have $$S_p=\frac{\pi ^2 p}{6}+\left(-\log(p)+\frac{\pi ^2}{6}-\gamma
   -1\right)-\frac{1}{p}+\frac{5}{12
   p^2}+O\left(\frac{1}{p^3}\right)$$ where $\gamma$ is Euler's constant.
Just to give you an idea, the exact summation for $n=10$ would lead to $\approx 14.118477$ while the asymptotic formula given above  would lead to $\approx 14.118641$.
