Infinite sums and products 4 (#13)

$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n^2}=\sum_{n=1}^{\infty}\frac{1}{n^2}-2\sum_{n=1}^{\infty}\frac{1}{(2n)^2}=$ $=\sum_{n=1}^{\infty}\frac{1}{n^2}-2\cdot\frac{1}{2^2}\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^2}$ It's a well-known fact (see Basel Problem ) that $\zeta(2)=\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}$ So the answer is $\frac{1}{2}\cdot \frac{\pi^2}{6}=\boxed{0.82...}$

Notice that if we add $\frac{2}{2^2}+\frac{2}{4^2}+\frac{2}{6^2}+\dots$ to the series,we get: $\left(1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\dots\right)+\left(\frac{2}{2^2}+\frac{2}{4^2}+\dots\right)=1+\frac{1}{2^2}+\frac{1}{3^2}+\dots$ It is a well known fact that $1+\frac{1}{2^2}+\frac{1}{3^2}+\dots=\frac{\pi^2}{6}$ .Also note that: $\frac{2}{2^2}+\frac{2}{4^2}+\dots=2\left(\frac{1}{2^2}+\frac{1}{4^2}+\dots\right)\\=2\times\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\dots\right)\\=\frac{1}{2}\times\frac{\pi^2}{6}=\frac{\pi^2}{12}$ Let $1-\frac{1}{2^2}+\frac{1}{3^2}+\dots=x$ then the abovementioned process can be written as: $x+\frac{\pi^2}{12}=\frac{\pi^2}{6}\\x=\frac{\pi^2}{6}-\frac{\pi^2}{12}=\frac{\pi^2}{12}=\boxed{0.82}\;to\;2\;d.p$

By the Weierstrass factorization theorem , we rewrite $\frac{\sin x}{x}$ in terms of its roots $x = n\pi$ for $n \in \mathbb{Z} \setminus \lbrace 0 \rbrace$ :

$\frac{\sin x}{x} = \Bigg(\prod_{m=1}^{\infty} (1+\frac{x}{m\pi})\Bigg) \Bigg(\prod_{n=1}^{\infty} (1-\frac{x}{n\pi})\Bigg) = \prod_{n=1}^{\infty} (1-\frac{x^{2}}{n^{2}\pi^{2}})$

Using Newton's sums to collect the $x^{2}$ terms, we see that the coefficient is $\displaystyle -\frac{1}{\pi^{2}} \sum_{k=1}^{\infty} \frac{1}{k^{2}}$ .

The Taylor series of $\sin x$ is $\displaystyle \sum_{k=1}^{\infty} \frac{(-1)^{k}x^{2k+1}}{(2k+1)!}$ . From this, we then know $\displaystyle \frac{\sin x}{x} = \sum_{k=1}^{\infty} \frac{(-1)^{k}x^{2k}}{(2k+1)!}$ .

The coefficient here of $x^{2}$ , then, is $-\frac{1}{6}$ . Equating the two coefficients and rearranging yields

$\sum_{k=1}^{\infty} \frac{1}{k^{2}} = \frac{\pi^{2}}{6}$

Multiplying both sides by $1-\frac{1}{2} = \frac{1}{2}$ , we get

$\sum_{k=1}^{\infty} \frac{1}{k^{2}}-\frac{2}{(2k)^{2}} = \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k^{2}} = \boxed{\frac{\pi^{2}}{12}}$

First of all, observe that

$\sum_{n=1}^{\infty} \frac{1}{(2n)^{2}} = \frac{1}{4} \zeta(2)$

and

$\sum_{n=1}^{\infty} \frac{1}{(2n-1)^{2}} = \zeta(2) - \sum_{n=1}^{\infty} \frac{1}{(2n)^{2}} = \zeta(2) - \frac{1}{4} \zeta(2) = \frac{3}{4} \zeta(2)$

Then also observe that the sum we want may be written as

$\sum_{n=1}^{\infty} \frac{1}{(2n-1)^{2}} - \sum_{n=1}^{\infty} \frac{1}{(2n)^{2}} = \frac{3}{4} \zeta(2) - \frac{1}{4} \zeta(2) = \frac{1}{2} \zeta(2) = \frac{\pi^{2}}{12} \approx 0.82$