euler and infinite series

[aeh49]

In his Introductio of 1748, Euler gave the sum of 1/1^n + 1/2^n + 1/3^n +... for even values of n from n=2 through n=26.
Formally show that pi-squared/12 = 1/1^2 - 1/2^2 + 1/3^2 - 1/4^2 + ....

Thank you

 

[galactus]

$\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^{2}}=1-\frac{1}{2^{2}}+\frac{1}{3^{2}}-\frac{1}{4^{2}}+.................$

Now, how in depth must you go?. Can we play off of other already known summations?. If not, we can try a more in-depth route.

We can use the well known $\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{{\pi}^{2}}{6}$

Break them up in to odds and evens:

$\displaystyle \frac{1}{2^{2}}+\frac{1}{4^{2}}+\frac{1}{6^{2}}+....=\sum_{k=1}^{\infty}\frac{1}{(2k)^{2}}=\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{1}{4}\cdot\frac{{\pi}^{2}}{6}=\frac{{\pi}^{2}}{24}$

Since these were all negatives, we can write $\displaystyle \frac{{-\pi}^{2}}{24}$

Now, the odds:

$\displaystyle 1+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}+........ = \sum_{k=0}^{\infty}\frac{1}{(2k+1)^{2}}=\frac{3}{4}\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{{\pi}^{2}}{8}$

$\displaystyle \frac{{\pi}^{2}}{8}-\frac{{\pi}^{2}}{24}=\frac{{\pi}^{2}}{12}$

I played off of already well-known identities. The actual derivations of these can be involved.

For instance, to find the latter summation, one can note that $\displaystyle \int_{0}^{1}\frac{sin^{-1}(t)}{\sqrt{1-t^{2}}}dt=\frac{{\pi}^{2}}{8}$

This, combined with various expansions and recursions will give us the result.