Can the Basel problem be solved by Leibniz today?

Solution 1:

Note that proving $$\sum_{k=1}^{\infty} \dfrac1{k^2} = \dfrac{\pi^2}6 \,\,\,\,\,\, (\spadesuit)$$ is equivalent to proving $$\sum_{k=0}^{\infty} \dfrac1{(2k+1)^2} = \dfrac{\pi^2}{8} \,\,\,\,\,\, (\clubsuit)$$ The hope for proving $(\clubsuit)$ instead of $(\spadesuit)$ is that squaring $\dfrac{\pi}4$ gives $\dfrac{\pi^2}{16}$ and adding this twice gives us $\dfrac{\pi^2}8$. We will in fact prove that $$\sum_{k=-\infty}^{\infty} \dfrac1{(2k+1)^2} = \left(\sum_{k=-\infty}^{\infty} \dfrac{(-1)^k}{(2k+1)} \right)^2$$ Since we know $$\sum_{k=0}^{\infty} \dfrac{(-1)^k}{(2k+1)} = \dfrac{\pi}4$$ and $$\sum_{k=0}^{N} \dfrac{(-1)^k}{(2k+1)} = \sum_{k=-(N+1)}^{-1} \dfrac{(-1)^k}{(2k+1)},$$ we have that $$\sum_{k=-\infty}^{\infty} \dfrac{(-1)^k}{(2k+1)} = \dfrac{\pi}2$$ Square the above to get \begin{align} \left(\sum_{k=-N-1}^{k=N} \dfrac{(-1)^k}{(2k+1)} \right)^2 & = \left(\sum_{k=-N-1}^{k=N} \dfrac{(-1)^k}{(2k+1)} \right) \left( \sum_{j=-N-1}^{j=N} \dfrac{(-1)^j}{(2j+1)} \right)\\ & = \sum_{k=-N-1}^{k=N} \sum_{j=-N-1}^{j=N} \dfrac{(-1)^{j+k}}{(2k+1)(2j+1)}\\ & = \sum_{\overset{j,k=-N-1}{j \neq k}}^{k=N} \dfrac{(-1)^{j+k}}{2(j-k)} \left(\dfrac1{(2k+1)} - \dfrac1{(2j+1)} \right) + \sum_{k=-N-1}^{k=N} \dfrac{(-1)^{2k}}{(2k+1)(2k+1)} \end{align} Hence, $$\left(\sum_{k=-N-1}^{k=N} \dfrac{(-1)^k}{(2k+1)} \right)^2 - \sum_{k=-N-1}^{k=N} \dfrac{1}{(2k+1)(2k+1)} = \sum_{\overset{j,k=-N-1}{j \neq k}}^{k=N} \dfrac{(-1)^{j+k}}{2(j-k)} \left(\dfrac1{(2k+1)} - \dfrac1{(2j+1)} \right)$$ Let us now show that $$\overbrace{\sum_{\overset{j,k=-N-1}{j \neq k}}^{k=N} \dfrac{(-1)^{j+k}}{2(j-k)} \left(\dfrac1{(2k+1)} - \dfrac1{(2j+1)} \right)}^{(\heartsuit)} \to 0 \text{ as } N \to \infty$$ We have \begin{align} (\heartsuit) & = \sum_{\overset{j,k=-N-1}{j \neq k}}^{k=N} \dfrac{(-1)^{j+k}}{2(j-k)} \dfrac1{(2k+1)} - \sum_{\overset{j,k=-N-1}{j \neq k}}^{k=N} \dfrac{(-1)^{j+k}}{2(j-k)} \dfrac1{(2j+1)}\\ & = \sum_{\overset{j,k=-N-1}{j \neq k}}^{k=N} \dfrac{(-1)^{j+k}}{2(j-k)} \dfrac1{(2k+1)} + \sum_{\overset{j,k=-N-1}{j \neq k}}^{k=N} \dfrac{(-1)^{j+k}}{2(k-j)} \dfrac1{(2j+1)}\\ & = 2 \times \left(\sum_{\overset{j,k=-N-1}{j \neq k}}^{k=N} \dfrac{(-1)^{j+k}}{2(j-k)} \dfrac1{(2k+1)} \right)\\ & = \sum_{\overset{j,k=-N-1}{j \neq k}}^{k=N} \dfrac{(-1)^{j+k}}{(j-k)} \dfrac1{(2k+1)} = \sum_{k=-N-1}^N \dfrac{(-1)^k}{(2k+1)} \underbrace{\left(\sum_{\overset{j=-N-1}{j \neq k}}^N \dfrac{(-1)^{j}}{(j-k)}\right)}_{(\diamondsuit_{k})} \end{align} Let us simplify $(\diamondsuit_{k})$ a bit. Assuming $k \neq -N-1$, we have \begin{align} \sum_{\overset{j=-N-1}{j \neq k}}^N \dfrac{(-1)^{j}}{(j-k)} & = \sum_{j=k+1}^N \dfrac{(-1)^{j}}{(j-k)} + \sum_{j=-N-1}^{k-1} \dfrac{(-1)^{j}}{(j-k)}\\ & = \left(\dfrac{(-1)^{k+1}}{1} + \dfrac{(-1)^{k+2}}{2} + \cdots + \dfrac{(-1)^{N}}{N-k}\right)\\ & + \left(\dfrac{(-1)^{k-1}}{(-1)} + \dfrac{(-1)^{k-2}}{(-2)} + \cdots + \dfrac{(-1)^{-N-1}}{-N-1-k}\right)\\ & = (-1)^{k+1} \sum_{j=N-\vert k \vert +1}^{N+\vert k \vert +1} \dfrac{(-1)^j}{j} \end{align} If $k = -N-1$, we have $$\sum_{\overset{j=-N-1}{j \neq k}}^N \dfrac{(-1)^{j}}{(j-k)} = \sum_{j=-N}^N \dfrac{(-1)^{j}}{(j+N+1)} = (-1)^{N-1} \sum_{j=1}^{2N+1} \dfrac{(-1)^j}{j}$$ We now have $$(\heartsuit) = \sum_{k=0}^N \dfrac{(-1)^k \diamondsuit_k + (-1)^{-k-1} \diamondsuit_{-k-1}}{2k+1} = \sum_{k=0}^N \dfrac{(-1)^k \left(\diamondsuit_k - \diamondsuit_{-k-1} \right)}{2k+1}$$ Now for $k \geq 0$ \begin{align} \left(\diamondsuit_k - \diamondsuit_{-k-1} \right) & = (-1)^{k+1} \sum_{j=N-k +1}^{N+k +1} \dfrac{(-1)^j}{j} - (-1)^{-k} \sum_{j=N-(k+1) +1}^{N+(k+1) +1} \dfrac{(-1)^j}{j}\\ & = 2 \cdot (-1)^{k+1} \cdot \sum_{j=N-k+1}^{N+ k +1} \dfrac{(-1)^j}{j} + (-1)^{N+1} \left( \dfrac1{N+k+2} + \dfrac1{N-k}\right) \end{align} Hence, $$\left \vert \diamondsuit_k - \diamondsuit_{-k-1} \right \vert = \mathcal{O} \left( \dfrac1N\right)$$ $$\left \vert (\heartsuit) \right \vert \leq \sum_{k=0}^N \dfrac1{2k+1} \mathcal{O}(1/N) = \mathcal{O}(\log(2N+1)/N) \to 0$$ Hence, $$\left(\sum_{k=-N-1}^{k=N} \dfrac{(-1)^k}{(2k+1)} \right)^2 - \sum_{k=-N-1}^{k=N} \dfrac{1}{(2k+1)(2k+1)} \to 0$$ Hence, $$\left(\sum_{k=-\infty}^{\infty} \dfrac{(-1)^k}{(2k+1)} \right)^2 = \sum_{k=-\infty}^{\infty} \dfrac{1}{(2k+1)(2k+1)} = 2 \sum_{k=0}^{\infty} \dfrac{1}{(2k+1)^2} = 2 \cdot \dfrac34 \cdot \zeta(2)$$ Hence,$$\boxed{\zeta(2) = \dfrac23 \cdot \dfrac{\pi^2}4 = \dfrac{\pi^2}6}$$

Solution 2:

The idea that Leibniz's series $$\frac \pi 4 = \sum_{i = 0}^\infty \frac{(-1)^i}{2i + 1} \tag{1}$$ could be used to prove Euler's $$\zeta(2) = \frac{\pi^2}{6} = \sum_{i=1}^\infty \frac{1}{i^2} \tag{2}$$ seems even more tantalizing when you consider that $(2)$ is equivalent to $$\frac{\pi^2}{8} = \sum_{i=0}^\infty \frac{1}{(2i+1)^2} \tag{3}.$$ The equivalence between $(2)$ and $(3)$ is clear from $$\frac{3}{4}\zeta(2) = \sum_{i=1}^\infty \frac{1}{i^2}- \sum_{i=1}^\infty \frac{1}{(2i)^2}= \sum_{i=0}^\infty \frac{1}{(2i+1)^2}.$$ Compare $(1)$ and $(3)$! My goodness.

Have you already looked at Different methods to compute $\sum\limits_{n=1}^\infty \frac{1}{n^2}$ and http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf ?

Solution 3:

One of the best way without leibnizSince $\int_0^1 \frac{dx}{1+x^2}=\frac{\pi}{4}$, we have

$$\frac{\pi^2}{16}=\int_0^1\int_0^1\frac{dydx}{(1+x^2)(1+y^2)}\overset{t=xy}{=}\int_0^1\int_0^x\frac{dtdx}{x(1+x^2)(1+t^2/x^2)}$$

$$=\frac12\int_0^1\int_t^1\frac{dxdt}{x(1+x^2)(1+t^2/x^2)}\overset{x^2\to x}{=}\frac12\int_0^1\left(\int_{t^2}^1\frac{dx}{(1+x)(x+t^2)}\right)dt$$

$$=-\frac12\int_0^1\frac{\ln\left(\frac{4t^2}{(1+t^2)^2}\right)}{1-t^2}dt\overset{t=\frac{1-x}{1+x}}{=}-\frac12\int_0^1\frac{\ln\left(\frac{1-x^2}{1+x^2}\right)}{x}dx$$

$$\overset{x^2\to x}{=}-\frac14\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)}{x}dx=-\frac14\int_0^1\frac{\ln\left(\frac{(1-x)^2}{1-x^2}\right)}{x}dx$$

$$=-\frac12\int_0^1\frac{\ln(1-x)}{x}dx+\frac14\underbrace{\int_0^1\frac{\ln(1-x^2)}{x}dx}_{x^2\to x}$$

$$=-\frac38\int_0^1\frac{\ln(1-x)}{x}dx\Longrightarrow \int_0^1\frac{-\ln(1-x)}{x}dx=\frac{\pi^2}{6}$$

Remark:

This solution can be considered a proof that $\zeta(2)=\frac{\pi^2}{6}$ as we have $\int_0^1\frac{-\ln(1-x)}{x}dx=\text{Li}_2(x)|_0^1=\text{Li}_2(1)=\zeta(2)$