On the rate of convergence of the Basel problem

Question

Prerequisites: $\quad \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}.$

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Now, I'm wondering if:

$$ \sum_{k=1}^{\infty} \left( \frac{\pi^2}{6} - \sum_{j=1}^{j=k} \frac{1}{j^2} \right) $$

converges or diverges.

From memory, I think that

$$ \lim_{k\to\infty}\ \frac{ \frac{\pi^2}{6} - \displaystyle\sum_{j=1}^{j=k} \frac{1}{j^2} } { \frac{\pi^2}{6} - \displaystyle\sum_{j=1}^{j={k+1}} \frac{1}{j^2} } = 1,$$

although I can't remember if I proved this or if it was true based on heuristic calculations with a calculator. If that limit is correct then we cannot use the ratio test to conclude anything about the series.

So is the value of the limit correct, and why? Does the series converge or diverge, and what is the proof for that also, please?

Answer 1

There is a simple way, by diagram, to estimate a sum. When the endpoints are finite all of them vary, but $\infty$ stays the same.

$$ \int_{k+1}^\infty \; \frac{1}{x^2} \; dx \; < \; \sum_{j=k+1}^\infty \; \frac{1}{j^2} \; < \; \int_{k}^\infty \; \frac{1}{x^2} \; dx $$

$$ \frac{1}{k+1} \; \; < \; \sum_{j=k+1}^\infty \; \frac{1}{j^2} \; < \; \frac{1}{k} $$

If we name the tail $T_k = \sum_{j=k+1}^\infty \; \frac{1}{j^2} $ we see that $\frac{1}{k+1} < T_k < \frac{1}{k}.$

$$ \int_1^{N+1} \; \frac{1}{x+1} \; dx \; < \; \sum_{k=1}^N \; \frac{1}{k+1} \; < \; \sum_{k=1}^N \; T_k \; $$ or

$$ \log(N+2) - \log 2 < \int_1^{N+1} \; \frac{1}{x+1} \; dx \; < \; \sum_{k=1}^N \; \frac{1}{k+1} \; < \; \sum_{k=1}^N \; T_k \; $$

Answer 2

$\begin{array}\\ s(k) &=\frac{\pi^2}{6} - \sum_{j=1}^{j=k} \frac{1}{j^2}\\ &=\sum_{j=k+1}^{\infty} \frac{1}{j^2}\\ &>\sum_{j=k+1}^{\infty} \frac{1}{j(j+1)}\\ &=\sum_{j=k+1}^{\infty} (\frac1{j}-\frac1{j+1})\\ &=\frac1{k+1}\\ \end{array} $

so $\sum_{k=1}^{\infty} s(k) $ diverges.

Also

$\begin{array}\\ s(k) &=\sum_{j=k+1}^{\infty} \frac{1}{j^2}\\ &<\sum_{j=k+1}^{\infty} \frac{1}{j(j-1)}\\ &=\sum_{j=k+1}^{\infty} (\frac1{j-1}-\frac1{j})\\ &=\frac1{k}\\ \end{array} $

so $\sum_{k=1}^{n} s(k) $ diverges like $\ln(n)$.

Answer 3

$$\sum_{k=1}^{\infty} \left( \frac{\pi^2}{6} - \sum_{j=1}^{j=k} \frac{1}{j^2} \right)=\sum_{k=1}^{\infty} \left( \frac{\pi^2}{6} -H_k^{(2)}\right)=\sum_{k=1}^{\infty}a_k$$

For large values of $k$, using the asymptotics of generalized harmonic numbers $$a_k=\frac{1}{k}-\frac{1}{2k^2}+O\left(\frac{1}{k^3}\right)\quad \implies \quad \text{divergence}$$

Concerning the partial sums $$S_n=\sum_{k=1}^{n}a_k=H_{n+1}-(n+1) H_{n+1}^{(2)}+\frac{\pi ^2 n}{6}$$ so, for large $n$ $$S_n=\left(1+\gamma -\frac{\pi ^2}{6}\right)+\log(n) +\frac{1}{n}-\frac{5}{12 n^2}+O\left(\frac{1}{n^3}\right)$$

For $n=1000$, the truncated series leads to $\color{red}{6.841036460}37$ while the exact value is $\color{red}{6.84103646054}$