An accurate, rapidly converging estimate of $\sum_{j = 1}^{\infty}\frac 1 {j^2}$ (the Basel Problem) using only elementary calculus

Question

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:

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

as this plot suggests:

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?

Answer 1

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.

Answer 2

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).