# Can the Basel problem be solved by Leibniz today?

It is well known that Leibniz derived the series \begin{align} \frac{\pi}{4}&=\sum_{i=0}^\infty \frac{(-1)^i}{2i+1},\tag{1} \end{align} but apparently he did not prove that \begin{align} \frac{\pi^2}{6}&=\sum_{i=1}^\infty \frac{1}{i^2}.\tag{2} \end{align} Euler did, in 1741 (unfortunately, after the demise of Leibniz). Note that this was also before the time of Fourier.

My question: do we now have the tools to prove (2) using solely (1) as the definition of $\pi$? Any positive/negative results would be much appreciated. Thanks!

Clarification: I am not looking for a full-fledged rigorous proof of (1)$\Rightarrow$(2). An estimate that (2) should hold, given (1), would qualify as an answer.

• It would be quite surprising if you could prove, or even give a heuristic argument, for (1) => (2). They are values of two different $L$-functions (namely $\zeta(s)$ and $L(s,\chi)$ for the nontrivial character modulo $4$) at two different values of $s$ ($s=1$ v. $s=2$.) – stopple Mar 8 '13 at 15:39

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,$$\boxed{\zeta(2) = \dfrac23 \cdot \dfrac{\pi^2}4 = \dfrac{\pi^2}6}$$
• This is a great answer! Quite $(\heartsuit)$ly. – sai Mar 9 '13 at 23:37
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.