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Let us denote by $S$ the sum of the series $\displaystyle\zeta(2)=1+\frac1{2^2}+\frac1{3^2}+\cdots$

Yes, I know (and you know) that $S=\frac{\pi^2}6$, but that is not relevant for the question that I am about to ask.

This series converges slowly. In fact, the sequence$$\left(S-\sum_{k=1}^n\frac1{k^2}\right)_{n\in\mathbb N}$$converges to $0$ at about the same rate as $\left(\frac1n\right)_{n\in\mathbb N}$. My question is about speeding up the rate of convergence of this series. More precisely, it is this: prove that there is a number $K\in(0,1]$ such that$$(\forall n\in\mathbb{N}):\left|S-\frac2{2n+1}-\sum_{k=1}^n\frac1{k^2}\right|\leqslant\frac K{n^3}.$$

Added note: Because of some comments that I got, I want to make this clear: I know an answer to this question.

José Carlos Santos
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  • One good technique for answering such questions is using the Euler–Maclaurin formula; see, in particular, the section on Asymptotic expansion of sums. After reading that, perhaps you can answer the question yourself! – Harald Hanche-Olsen Jun 22 '17 at 10:25
  • I know an answer. And it does not involve the Euler–Maclaurin formula. – José Carlos Santos Jun 22 '17 at 10:27
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    Okay; so if you know an answer, why do you ask for one? To see if someone has a better answer? But then it might be good if you outline your own technique, so nobody wastes time duplicating it. – Harald Hanche-Olsen Jun 22 '17 at 10:29
  • @HaraldHanche-Olsen I will be delighted if someone provides a better answer than the one that I have in mind. But I am sorry if my way of asking the question was innapropriate. It is only the third time that I post a question here. – José Carlos Santos Jun 22 '17 at 10:31
  • Hint : look at $2\int_1^{n+1} \lfloor x \rfloor x^{-3}dx$ and $2\int_1^\infty \lfloor x \rfloor x^{-3}dx = \zeta(2)$ (see the Abel summation formula, alternatively use summation by parts) also note $x-\lfloor x \rfloor$ is periodic and bounded. – reuns Jun 22 '17 at 10:42
  • It's not inappropriate, but I guess it would be better if you wrote so in your question, clearly: that you've got a proof, but you hope somebody can find a better one (shorter, better constant,...). Would you like $K=1/12,$ divided by $(n+1/2)^3,$ instead? –  Jun 22 '17 at 10:43
  • @ProfessorVector If you have a proof of the fact that the last inequality holds with $\frac K{n^3}$ replaced by $\frac1{12(n+1/2)^3}$, that will be fine by me. By the way, the best $K$ that I got is $1/9$. – José Carlos Santos Jun 22 '17 at 11:00
  • Would this series help? $$\pi^2 = 10-\sum_{k=0}^\infty \frac{1}{((k+1)(k+2))^3}$$ – Jaume Oliver Lafont Jun 22 '17 at 11:03
  • @JaumeOliverLafont I don't see how. – José Carlos Santos Jun 22 '17 at 11:05
  • I tried to develop in an answer. – Jaume Oliver Lafont Jun 22 '17 at 12:22
  • I think that a companion question of yours (that is if you want to study it in your home) is what about different Dirichlet series. I am saying that maybe can be interesting to study your question but now for different examples of Dirichlet series like next, $\sum_{n=1}^\infty\frac{\mu(n)}{n^2}$, where $\mu(n)$ denotes the Möbius function, or assumming that we know $\zeta(2)=\frac{\pi^2}{6}$ for example this $\sum_{n=1}^\infty\frac{\sigma(n)}{n^3}$, where $\sigma(n)=\sum_{d\mid n}d$. I hope don't disturb with my comment, good week. –  Feb 25 '18 at 11:33

5 Answers5

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Obviously, the rest after the $n$th partial sum is $$S-\sum_{k=1}^n\frac1{k^2}=\sum_{k=n+1}^\infty\frac1{k^2}.$$ Let's approximate that with some similar series with known partial sums, so a telescoping series would be nice. A convenient choice would be $$\frac1{k^2-1/4}=\frac1{k-1/2}-\frac1{k+1/2},$$ so $$\sum_{k=n+1}^\infty\frac1{k^2-1/4}=\frac1{n+1/2}=\frac2{2n+1}$$ is the main part. We're left with an estimate for the error, i.e. for $$\sum_{k=n+1}^\infty\left(\frac1{k^2-1/4}-\frac1{k^2}\right)=\sum_{k=n+1}^\infty\frac{1/4}{k^2(k^2-1/4)}.$$ Let's try with some telescoping series, too: We have $$\frac1{(k-1/2)^3}-\frac1{(k+1/2)^3}=\frac{3k^2+1/4}{(k^2-1/4)^3}\ge\frac{3}{(k^2-1/4)^2}\ge12\cdot\frac{1/4}{k^2(k^2-1/4)},$$ and this means $$\sum_{k=n+1}^\infty\frac{1/4}{k^2(k^2-1/4)}\le\frac{1/12}{(n+1/2)^3}.$$

  • Nice! That's basically the answer that I had in mind. – José Carlos Santos Jun 22 '17 at 11:14
  • The estimate is sharp, in fact, you can use it as the next term for convergence acceleration: $$\sum^n_{k=1}\frac1{k^2}+\frac1{n+1/2}-\frac1{12(n+1/2)^3}$$ with $n=400$ is good for 14 decimals. But: To compute the Zeta function (for positive real arguments), I use the corresponding alternating series, where far better methods of acceleration (and certainly better than the Van Wijngaarden transformation) are available. –  Jun 22 '17 at 16:47
  • For most purposes the formula to use is this series computed using this algorithm @JoséCarlosSantos – reuns Jun 22 '17 at 19:39
  • @user1952009 Well, I prefer Borwein's method, because it works for very many alternating series (those with totally monotonic summands) and is essentially optimal, concerning convergence rate. –  Jun 22 '17 at 20:19
  • A priori the Euler transform is much easier to prove and implement. – reuns Jun 22 '17 at 20:56
  • @Professor vector you are funny lady – gymbvghjkgkjkhgfkl Jun 24 '17 at 22:30
  • @professor vector this link may be very useful so learn from it https://en.m.wikipedia.org/wiki/The_finger – gymbvghjkgkjkhgfkl Jun 24 '17 at 22:33
  • @NightOfPower You seem to be a bit funny yourself. Did you really think the professor of arithmancy at Hogwarts is a lady? Rowling is a writer, she's entitled to some poetic license! –  Jun 25 '17 at 05:20
  • @professor vector can't understand what you are writing but anyway you are ****! Lol – gymbvghjkgkjkhgfkl Jun 25 '17 at 18:46
  • @NightOfPower I think "I can't understand" is ok (and it's certainly honest), even though I don't know very many people who would be proud of it. –  Jun 25 '17 at 20:15
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Multiplying $\dfrac{1}{n^2}$ by $\dfrac{3^n-1}{\binom{2n}{n}}$ we can speed up convergence . That is: $$\zeta(2)=\displaystyle\sum_{n=1}^\infty \frac{3^n-1}{n^2\binom{2n}{n}}$$

Pedja
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Set \begin{align} a_m &= \sum_{m=1}^{m}\frac{1}{n^2}\\ b_m &= 2 \sum_{n=1}^{m}\frac{(-1)^{n+1}}{n^2}, \qquad m>1 \end{align} Then perhaps prove or disprove $$\bigg |\frac{\zeta(2)-b_n}{\zeta(2)-a_n} \bigg | \rightarrow 1$$

Whcih would imply $b_m$ converges a lot quicker than $a_m$.

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From series

$$\pi^2 = 10 - \sum_{k=0}^\infty \frac{1}{((k+1)(k+2))^3}$$

we can write

$$\frac{\pi^2}{6}=\frac{5}{3}-\frac{1}{6}\sum_{k=0}^\infty \frac{1}{((k+1)(k+2))^3}$$

Equivalently,

$$\begin{align} \frac{5}{3}-\frac{1}{6}\sum_{k=0}^{n} \frac{1}{((k+1)(k+2))^3} -\frac{\pi^2}{6}&= \frac{1}{6}\sum_{k=n+1}^\infty \frac{1}{((k+1)(k+2))^3}\\ &< \frac{1}{6}\sum_{k=n+1}^\infty \frac{1}{(k+1)^6}\\ &< \frac{1}{6}\frac{1}{5n^5}=\frac{1}{30n^5}\\ \end{align}$$

Jaume Oliver Lafont
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So, here is my own answer.

Since$$S-\sum_{k=1}^n\frac1{k^2}=\sum_{k=n+1}^\infty\frac1{k^2}\text{ and }\frac2{2n+1}=\sum_{k=n+1}^\infty\left(\frac2{2k-1}-\frac2{2k+1}\right)=\sum_{k=n+1}^\infty\frac4{4k^2-1}\text,$$we know that\begin{align*}\frac2{2n+1}-S+\sum_{k=1}^n\frac1{k^2}&=\sum_{k=n+1}^\infty\left(\frac4{4k^2-1}-\frac1{k^2}\right)\\&=\sum_{k=n+1}^\infty\left(\frac1{4k^4-k^2}\right).\end{align*}So\begin{align*}0&<-S+\frac2{2n+1}+\sum_{k=1}^n\frac1{k^2}\\&<\int_n^{+\infty}\frac{dx}{4x^4-x^2}\\&<\int_n^{+\infty}\frac{dx}{3x^4}\\&=\frac1{9n^3}.\end{align*}So, it starts like the answer provided by Professor Vector, but the method used to find an upper bound is different. I think that Professor Vector's answer is better than mine because it leads to a sharper estimate and because it doesn't require integrals.

José Carlos Santos
  • 440,053