In his gorgeous paper “How to compute $\sum \frac{1}{n^2}$ by solving triangles”, Mikael Passare offers this idea for proving $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$:

Proof of equality of square and curved areas is based on another picture:

Recapitulation of Passare’s proof using formulas is as follows:

$$\sum_{n=1}^\infty \frac{1}{n^2} = \sum_{n=1}^\infty \int_0^\infty \frac{e^{-nx}}{n}\; dx\; = -\int_0^\infty \log(1-e^{-x})\; dx\; = \frac{\pi^2}{6}$$

There is also another paper dealing with geometric proof of $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$, in an entirely different way.

I tried to find a similar way to prove:

$$\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$$

but didn’t succeed. Maybe you will?

**Answer**

The first part is similar.

$$\dfrac{1}{n^4} = \dfrac{1}{n} \int_0^\infty \int_0^\infty \int_0^\infty e^{-n(x+y+z)}\; dx\; dy\; dz $$

so

$$\sum_{n=1}^\infty \dfrac{1}{n^4} = \int_0^\infty \int_0^\infty \int_0^\infty -\log(1 – e^{-(x+y+z)})\; dx\; dy\; dz$$

Now we’re integrating over an octant of $\mathbb R^3$. Change variables to $u = x$, $v = x+y$, $w = x+y+z$, with $du\; dv\; dw = dx\; dy\; dz$:

$$ \eqalign{\sum_{n=1}^\infty \dfrac{1}{n^4} &= \int_{w=0}^\infty \int_{v=0}^w \int_{u=0}^v -\log(1 – e^{-w})\; du\; dv\; dw\cr

&= -\int_0^\infty \dfrac{w^2 \log(1-e^{-w})}{2}\; dw\cr

} $$

The tricky part is evaluating that integral.

**Attribution***Source : Link , Question Author : VividD , Answer Author : Robert Israel*