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Although the Basel Problem $$\zeta(2) = \sum_{j = 1}^{\infty}\frac 1 {j^2}$$ took a hundred years to solve analytically, using elementary calculus we can easily get strikingly accurate bounds: $$\zeta(2) = \sum_{j = 1}^{n}\frac 1 {j^2} + \frac {t(n)}{n} + \frac{1 - t(n)}{n+1}$$ (for all $n$) for some $$0 < t(n) < 1.$$

Proof: This follows from $$\sum_{j = n}^{\infty}\frac 1 {j^2} = \int_n^\infty \frac 1 {\lfloor x \rfloor^2} \quad dx.$$

This formula gives at the midpoint ($t = \frac 1 2$) a very accurate estimate, with $< 0.23\%$ relative error after just a few terms ($n = 4$). It seems to be both simpler and more accurate than some of the other elementary approaches tried:

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What's striking is that $t$ clearly seems to approach $1/2$ as $n \to \infty$:

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as this plot suggests: enter image description here

Question

  • Can we prove (or even just explain intuitively) that as $n \to \infty, t(n) \to 1/2$?
  • Can we derive a formula or estimate for $t(n)$ (such as e.g. $t(n) = \frac 1 2 - \frac 1 {n^{6/5}} + ...$)?
  • Can this be used to solve the Basel Problem?
SRobertJames
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  • take a look at this post for the fun of it: https://math.stackexchange.com/questions/8337/different-ways-to-prove-sum-k-1-infty-frac1k2-frac-pi26-the-b – Math-fun Jan 23 '24 at 18:04
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    Look at this paper https://royalsocietypublishing.org/doi/10.1098/rspa.2017.0363. You are looking for the special case of https://en.wikipedia.org/wiki/Hurwitz_zeta_function for $a = n$ and $s = 2$ – Kroki Jan 23 '24 at 18:07
  • Try Euler–Maclaurin, but there are more rapidly converging options than that. Would you be satisfied with estimates of $\eta(2)=\frac12\zeta(2)$? It's an alternating series one can accelerate with $k$ iterations of the Shanks transformation so the first $n+2k$ terms' sum has a $O(1/n^{2k+1})$ error term. – J.G. Jan 23 '24 at 18:59
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    Cf. https://en.wikipedia.org/wiki/Kummer%27s_transformation_of_series – Travis Willse Jan 23 '24 at 23:54
  • @TravisWillse which is a fancy way of adding and subtracting $\gamma B$ but effectively. Pretty clever. – Kamal Saleh Jan 24 '24 at 01:00
  • @TravisWillse Can you elaborate - How do we apply Kummer's transformation here? What are $a, b, \gamma, B$? – SRobertJames Jan 24 '24 at 01:22
  • @SRobertJames I only mean to point out that Kummer's transformation has a similar aim to your method here. Since $\sum_{n = 1}^\infty \frac{1}{n (n + 1)} = 1$, we have $\sum_{n = 1}^\infty \frac{1}{n^2} = 1 + \sum_{n = 1}^\infty \frac{1}{n^2 (n + 1)}$. The $N$th partial sum of the latter series has remainder $\psi(1, N + 1) - \frac1{N + 1} \sim \frac{1}{2} N^{-2}$. – Travis Willse Jan 24 '24 at 02:04

2 Answers2

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From the way you introduce $t(n)$, it’s given by

\begin{eqnarray*} t(n) &=& \frac{\int_{n+1}^\infty\frac1{\lfloor x\rfloor^2}\mathrm dx-\int_{n+1}^\infty\frac1{x^2}\mathrm dx}{\int_{n+1}^\infty\frac1{(x-1)^2}\mathrm dx-\int_{n+1}^\infty\frac1{x^2}\mathrm dx} \\[8pt] &=& \frac{\int_{n+1}^\infty\frac1{\lfloor x\rfloor^2}\mathrm dx-\frac1{n+1}}{\frac1n-\frac1{n+1}}\;. \end{eqnarray*}

We can approximate the integral by writing it as a sum of integrals over unit intervals and expanding the integrands around the midpoints of the intervals:

\begin{eqnarray*} \int_{n+1}^\infty\frac1{\lfloor x\rfloor^2}\mathrm dx &=& \sum_{k=n+1}^\infty\int_k^{k+1}\frac1{k^2}\mathrm dx \\ &=& \sum_{k=n+1}^\infty\int_{k-\frac12}^{k+\frac12}\frac1{k^2}\mathrm du \\ &=& \sum_{k=n+1}^\infty\int_{k-\frac12}^{k+\frac12}\frac1{(u+k-u)^2}\mathrm du \\ &=& \sum_{k=n+1}^\infty\int_{k-\frac12}^{k+\frac12}\frac1{u^2}\cdot\frac1{\left(1+\frac{k-u}u\right)^2}\mathrm du \\ &=& \sum_{k=n+1}^\infty\int_{k-\frac12}^{k+\frac12}\frac1{u^2}\left(1-2\frac{k-u}u+3\left(\frac{k-u}u\right)^2+O\left(\left(\frac{k-u}u\right)^3\right)\right)\mathrm du \\ &=& \frac1{n+\frac12}-8\sum_{k=n+1}^\infty\frac1{\left(4k^2-1\right)^2}+4\sum_{k=n+1}^\infty\frac{4k^2+3}{\left(4k^2-1\right)^3}+\sum_{k=n+1}^\infty O\left(\frac1{k^5}\right) \\ &=& \frac1{n+\frac12}-\frac12\sum_{k=n+1}^\infty\frac1{k^4}+\frac14\sum_{k=n+1}^\infty\frac1{k^4}+\sum_{k=n+1}^\infty O\left(\frac1{k^5}\right) \\ &=& \frac1{n+\frac12}-\frac1{12}\frac1{n^3}+ O\left(\frac1{n^4}\right)\;. \end{eqnarray*}

Then

\begin{eqnarray*} t(n) &=& \frac{\frac1{n+\frac12}-\frac1{12}\frac1{n^3}+ O\left(\frac1{n^4}\right)-\frac1{n+1}}{\frac1n-\frac1{n+1}} \\[2pt] &=& \frac{\frac1{n+\frac12}-\frac1{n+1}}{\frac1n-\frac1{n+1}}+\left(n^2+O(n)\right)\left(-\frac1{12}\frac1{n^3}+O\left(\frac1{n^4}\right)\right) \\[4pt] &=& \frac12-\frac14\cdot\frac1n-\frac1{12}\cdot\frac1n+O\left(\frac1{n^2}\right) \\[4pt] &=& \frac12-\frac13\cdot\frac1n+O\left(\frac1{n^2}\right)\;, \end{eqnarray*}

in excellent agreement with your numerical values.

joriki
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  • Thanks! Can you explain the last few lines of the first display? How did you simplify, then remove, the infinite sum? – SRobertJames Jan 23 '24 at 22:39
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    @SRobertJames: Since I was aiming for an error of order $\frac1{n^4}$, I only retained the leading terms. You can expand the summands in inverse powers of $k$; the omitted terms are of order $\frac1{k^5}$ and thus sum to a term of order $\frac1{n^4}$. Likewise, the sum of $\frac1{k^4}$ is $\frac13\cdot\frac1{n^3}$ plus a term of order $\frac1{n^4}$ (which could be proved by replacing it by an integral and bounding the error). All of these expansions can be carried to higher orders if you want to obtain the $\frac1{n^2}$ term in $t(n)$. – joriki Jan 23 '24 at 22:48
  • I see. If I wanted to repeat this derivation (to understand it better), would you recommend doing the last few lines on my own, or using software such as Wolfram? For example, the line which removes the integral seems straightforward but very laborious (and hard to do without error). – SRobertJames Jan 23 '24 at 22:52
  • @SRobertJames: I did a lot of it with Wolfram|Alpha – the two integrals and the expansion of $\frac{\frac1{n+\frac12}-\frac1{n+1}}{\frac1n-\frac1{n+1}}$. Life is too short to do things that computers can do :-) The main thing here to sharpen your understanding is to understand how to control the orders of approximation. For instance, note that I had to take along another term in the integral even though the first one already had the right power of $k$ because the first one with $k-u$ has cancellation that the second one with $(k-u)^2$ doesn't have. – joriki Jan 23 '24 at 23:01
  • If you want to go to the next order, you don't need any additional terms in the integral because the next one with $(k-u)^3$ again has cancellation and yields a summand of order $\frac1{k^6}$ that you don't need to get the $\frac1{n^2}$ term in $t(n)$. – joriki Jan 23 '24 at 23:01
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I can answer the second point, although it might not be satisfactory. Solving for $t(n)$: $$t(n)=n(n+1)\left(\zeta(2)-\sum_{k=1}^n\frac1{k^2}\right)-n$$So we can rewrite the first point as proving $$\lim_{n\rightarrow\infty}\sum_{k=1}^\infty\frac{n^2+n}{(k+n)^2(2n+1)}=\frac12$$Thanks to Semiclassical's comment, we could simplify this to $$\lim_{n\rightarrow\infty}\sum_{k=1}^\infty\frac n{(k+n)^2}=1$$Which can be equated to the improper integral $\int_0^\infty(1+x)^{-2}dx$ (Thanks to @J.G., I forgot to mention you). This is because we can use the fact that $$\int_a^bf(x)dx=\lim_{n\rightarrow\infty}\frac{b-a}n\sum_{k=1}^nf\left(a+\frac{k(b-a)}{n}\right)$$And summing this from $i=0$ to infinity with $a=i$ and $b=i+1$ (here $f(x)=(1+x)^{-2}$. prove it).

Kamal Saleh
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