What does the zeta squared function converge to

Question

Define a series: $f(n) = \frac{1}{n^2}$ where $n$ is a natural number. This series converges to $0$.

Now, consider the series $S(n) = \sum\limits_{i=0}^n f(i)$. This is the Reimann Zeta function evaluated at $2$ and it converges to $\frac{\pi^2}{6}$.

Now consider the series $R(n) = \sum\limits_{i=0}^n S(i)$. Does this converge to anything? If so, what does it converge to?

And if $R(n)$ diverges, what about $R_k(n)$ where we start with $f(n) = \frac{1}{n^k}$ instead? At what value of $k$ will we get convergence?

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My attempt:

I managed to establish that

$$R(n) = \sum\limits_{i=1}^n \frac{(n-i+1)}{i^2} = n \sum\limits_{i=1}^n \frac{1}{i^2}-\sum\limits_{i=1}^n\frac{1}{i}+\sum\limits_{i=1}^n\frac{1}{i^2} = n \sum\limits_{i=1}^n \frac{1}{i^2}-\sum\limits_{i=1}^n\frac{1}{i}+\frac{\pi^2}{6}$$

With the first two terms, we get an $\infty - \infty$ form which is indeterminate. It seems like the first sum should overpower the second one. This is because the first one is linear is $n$ while the second one is logarithmic. Not sure if this argument is precise enough. And now what about the $k$ for which $R_k(n)$ converges?

Answer 1

Your series doesn't exist. This is because the summand approaches $\frac{\pi^2}6$ and not $0$. Same thing for all $k$, the summand of $R_k$ will approach $\zeta(k)>1$.

Answer 2

Using generalized harmonic numbers $$S_2(n) = \sum\limits_{i=1}^n f(i)=H_n^{(2)}$$ Using the polygamma function $$R_2(n) = \sum\limits_{i=0}^n S_2(i)=(n+1) H_{n+1}^{(2)}-\psi ^{(0)}(n+2)-\gamma$$

For large values of $n$ $$R_2(n)=\frac{\pi ^2 }{6}n-\log(n)+\left(\frac{\pi ^2}{6}-\gamma -1\right)-\frac{1}{n}+\frac{5}{12 n^2}+O\left(\frac{1}{n^3}\right)$$

For the next one $$S_3(n) = H_n^{(3)}$$ $$R_3(n) =(n+1) H_{n+1}^{(3)}+\psi ^{(1)}(n+2)-\frac{\pi ^2}{6}$$ $$R_3(n) = \zeta (3)n+\left(\zeta (3)-\frac{\pi ^2}{6}\right)+\frac{1}{2 n}-\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)$$