# Visual proof of $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$?

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?

• What an ingenious proof in that paper! Commented May 15, 2015 at 23:02
• You might want to check the following link which discusses the methods solving this kind of problems. Some of them can be generalized to your case=) math.stackexchange.com/questions/8337/… Commented May 15, 2015 at 23:16
• Beautiful. It would be marvelous if "geometric" proofs of this kind could be found for all values of $\zeta (2k)$. Commented May 29, 2015 at 16:12
• How is this proof visual? I "see" neither that the curved area equals $\pi^4/90$ nor that the areas of the squares equal the areas of the exponentials. (I understand the proof, I don't understand what the pictures are good for) Commented Apr 12, 2016 at 14:04
• @GrumpyParsnip, I found another web location and corrected the link. Commented Apr 18, 2017 at 21:03

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.

• The integral is indeed tricky. There is at least one way to do it which is to say that it is equal to $\zeta(4)=\frac{\pi^4}{90}$ which is but circular. :-) Commented May 30, 2015 at 4:44
• The integral may be computed as follows: first integrate by parts once with $u= \log(1-e^{-w})$, $v' = w^2/2$. The boundary terms vanish and one finds: $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{1}{6} \int_0^\infty \frac{w^3}{\exp(w)-1} \mathrm{d}w$. The remaining integral may be computed by choice of an appropriate contour as in the second answer to this question: math.stackexchange.com/questions/99843/…. One finds that the integral gives $\pi/15$, hence the final result is that $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi}{90} \square$. Commented Apr 12, 2016 at 14:39
• @Cyclone Can you add this comment as a separate answer, with some more details, and step-by-step explanations? Commented Apr 13, 2016 at 10:50
• @RobertIsrael, yesterday was your best day on this site, 490 points, congratulations. Commented Apr 14, 2016 at 10:14
• @ RobertIsrael The integral representation of the sum can be found more easily (and more generally) by just following Bernhard Riemann (1859): let $n^{-s} = \frac{1}{\Gamma(s)} \int_0^\infty t^{s-1} e^{-n t}\,dt$ and do the geometric sum under the integral giving finally $\sum_{n=1}^\infty \frac{1}{n^s} = \frac{1}{\Gamma(s)} \int_0^\infty \frac{t^{s-1}}{ e^{ t}-1}\,dt$. Commented Jun 21, 2019 at 18:29

As requested by the OP, this is a detailed answer on how to calculate the integral in the final equality of Robert Israel's post, which is $$\sum\limits_{n=1}^\infty \frac{1}{n^4} = -\int_0^\infty \frac{w^2}{2}\log(1-e^{-w})\mathrm{d}w.$$ First integrate by parts with $u=\log(1-e^{-w})$, $v'=w^2/2$. This gives $$\phantom{texttext} \sum\limits_{n=1}^\infty \frac{1}{n^4} = -\underbrace{\frac{1}{6} w^3 \log(1-e^{-w})|_0^\infty}_{=0} + \frac{1}{6} \underbrace{\int_0^\infty \frac{w^3}{e^w-1} \mathrm{d}w}_{\equiv J}.\phantom{texttext} (1)$$ Now the remaining task is to compute the integral $J$. .............................................................................................................................................................................. This can be done by contour integration as explained in the second answer to the question "Contour integral for $x^3/(e^x-1)$?" (hereafter OA) . In order for the present answer to be as self-contained as possible, I will restate the answer given in the thread I linked (I give all credit to the author of the OA, the following derivation closely follows his and I recommend you read the OA). If it is not acceptable to restate an existing answer, I will remove this part.

Step 1: Choose the following contour, let's call it $\Gamma$, (figure source: OA) $\phantom{texexexexexexxt}$ Step 2: Consider the integral $$\oint_\Gamma \frac{z^4}{e^z-1} \mathrm{d}z,$$ which vanishes by Cauchy's theorem because $\Gamma$ is closed and the integrand is analytic. Writing out the contributions of the four edges and the two circle segments,

$$\int_\epsilon ^R \frac{x^4}{e^x - 1}\mathrm{d}x + \int_0 ^{2\pi} \frac{(R+iy)^4}{e^{R+iy}- 1}i\mathrm{d}y + \int_{R}^\epsilon \frac{(x+i2\pi)^4}{e^{x+i2\pi}-1}\mathrm{d}x + \int_0 ^{-\frac{\pi}{2}}\frac{(2 \pi i + \epsilon e^{i\theta})^4}{e^{2\pi i + \epsilon e^{i\theta}}-1} i\epsilon e^{i\theta}\mathrm{d}\theta + \int_{2\pi- \epsilon}^{\epsilon}\frac{(iy)^4}{e^{iy}-1}i\mathrm{d}y+ \int_{\frac{\pi}{2}}^{0}\frac{(\epsilon e^{i\theta})^4}{e^{\epsilon e^{i\theta}}-1}i\epsilon e^{i\theta} \mathrm{d}\theta = 0.$$

Step 3: Take the limit $R\to\infty$ and $\epsilon\to 0$. This eliminates the second and last terms respectively. Care must be taken in the 4th integral, because $\epsilon\to 0$ cannot be taken straightforwardly. Rather, use $\lim_{x\to0}x/(e^{ax}-1)=1/a$ to find that the 4th integral becomes $-8i\pi^5$. Next expand the power in the third term and note that the $x^4$ term cancels with the first integral. Splitting the 5th integral into real and imaginary parts leaves us with the equation $$-i8\pi \int_0 ^\infty \frac{x^3}{e^x - 1} \mathrm{d}x + 24\pi^2\int_0 ^\infty \frac{x^2}{e^x -1}\mathrm{d}x + i 32 \pi^3 \int_0 ^\infty \frac{x}{e^x - 1}\mathrm{d}x- 16\pi^4\int_0 ^\infty \frac{1}{e^x - 1}\mathrm{d}x -i8\pi^5+\frac{i}{2} \int_0 ^{2\pi} y^4 \mathrm{d}y - \frac{1}{2} \int_0 ^{2\pi} \frac{y^4 \sin y}{1-\cos y}\mathrm{d}y=0.$$

Step 4: Take the imaginary part to find $$-8 \pi J + 32 \pi^3 \int_0 ^\infty \frac{x}{e^x - 1}\mathrm{d}x - 8\pi^5+\frac{16\pi^5}{5} = 0$$ The remaining integral is shown, by the classic trick in the accepted answer to the linked question to be $$\int_0^\infty \frac{x}{e^x-1} \mathrm{d}x = \zeta(2).$$ The value of $\zeta(2)=\pi^2/6$ has been given in the question (it is incorrectly stated in the OA as $\pi^2/12$. This gives the final result $$J = \frac{\pi^4}{15},$$ which, by (1), completes the proof that $\zeta(4)=\pi^4/90 \text{ }\square$.