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

enter image description here

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

enter image description here

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?

  • 5
    $\begingroup$ What an ingenious proof in that paper! $\endgroup$ – Théophile May 15 '15 at 23:02
  • 1
    $\begingroup$ 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/… $\endgroup$ – Aprilius May 15 '15 at 23:16
  • $\begingroup$ the only visualization I can think of is in 4 dimensions, finishing Robert Israel's proof. Is that okay? I could try to project the darn thing myself. $\endgroup$ – cactus314 May 27 '15 at 18:57
  • 1
    $\begingroup$ Beautiful. It would be marvelous if "geometric" proofs of this kind could be found for all values of $\zeta (2k)$. $\endgroup$ – user_of_math May 29 '15 at 16:12
  • 1
    $\begingroup$ @GrumpyParsnip, I found another web location and corrected the link. $\endgroup$ – VividD Apr 18 '17 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.

  • $\begingroup$ 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. :-) $\endgroup$ – marwalix May 30 '15 at 4:44
  • 2
    $\begingroup$ 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 $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi}{90} \square$. $\endgroup$ – Cyclone Apr 12 '16 at 14:39
  • $\begingroup$ @Cyclone Can you add this comment as a separate answer, with some more details, and step-by-step explanations? $\endgroup$ – VividD Apr 13 '16 at 10:50
  • $\begingroup$ @RobertIsrael, yesterday was your best day on this site, 490 points, congratulations. $\endgroup$ – VividD Apr 14 '16 at 10:14
  • $\begingroup$ @ 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 $\sum_{n=1}^\infty \frac{1}{n^s} = \frac{1}{\Gamma(s)} \int_0^\infty \frac{t^{s-1}}{ e^{ t}-1}\,dt$. $\endgroup$ – Dr. Wolfgang Hintze Jun 21 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) enter image description here $\phantom{texexexexexexxt}$ 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 $\zeta(4)=\pi^4/90 \text{ }\square$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.