Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to prove this ? $$-\frac\pi2 = \lim_{x\to\infty}\sum_{n=1}^{\infty}(-1)^n \frac{x^{2n-1}}{(2n)! \ln 2n}$$

share|cite|improve this question
I tried to typeset your equation more nicely. Please check if I did right. – martini Sep 4 '12 at 16:14
What is the source? – Théophile Sep 4 '12 at 16:26
@ThomasAndrews: it looks rather straightforward to guess which infinity mick has in mind... – Fabian Sep 4 '12 at 16:34
Unless I'm missing something here, the expression $\,u\to\infty\,$ is always understood as "$\,u\,$ going to (plus) infinity", otherwise it must specifically be added a minus sign: $\,x\to -\infty\,$ – DonAntonio Sep 4 '12 at 16:37
As often when you see x^a/log(x)^b its related to number theory. Although i did not give it that tag. It appeared to me in an attempt to prove the prime twins conjecture. It reappeared when trying to prove RH. ( related to spacing of zero's ). – mick Sep 4 '12 at 18:08
up vote 20 down vote accepted

Using $$\int_0^\infty \left(2 n\right)^{-t} \mathrm{d} t = \frac{1}{\ln(2n)}$$ the sum becomes $$ \sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)! \cdot \ln(2n)} = \int_0^\infty \left(\sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)! \cdot (2n)^t} \right)\mathrm{d}t $$ Now, further using $$ \int_0^\infty u^{t-1} \mathrm{e}^{-2 n u} \mathrm{d} u = \Gamma(t) (2n)^{-t} $$ we rewrite the sum as a double integral: $$ \sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)! \cdot \ln(2n)} = \int_0^\infty \left( \int_0^\infty \frac{u^{t-1}}{\Gamma(t)}\mathrm{d}t \right) \frac{\cos\left(x \mathrm{e}^{-u}\right)-1}{x} \mathrm{d} u $$ In the large $x$ limit, the main contribution to the integral comes from large $u$. For large $u$, $$ \int_0^\infty \frac{u^{t-1}}{\Gamma(t)}\mathrm{d}t \approx \sum_{t=1}^\infty \frac{u^{t-1}}{\Gamma(t)} = \mathrm{e}^{u} $$ enter image description here

Thus: $$ \begin{eqnarray} \lim_{x \to \infty} \sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)! \cdot \ln(2n)} &=& \lim_{x \to \infty} \int_0^\infty \mathrm{e}^{u} \frac{\cos\left(x \mathrm{e}^{-u}\right)-1}{x} \mathrm{d} u = \lim_{x \to \infty} \int_1^\infty \frac{\cos\left(x/w\right)-1}{x} \mathrm{d} w \\ &=& \lim_{x \to \infty} \int_{1/x}^\infty \left(\cos\left(\frac{1}{v}\right)-1\right) \mathrm{d} v = \int_{0}^\infty \left(\cos\left(\frac{1}{v}\right)-1\right) \mathrm{d} v \\ &=& -\frac{\pi}{2} \end{eqnarray} $$

share|cite|improve this answer
Euh i did not get all those steps. It would be helpful if you explained what substitutions or partials you used. Also a plot is not a proof although i guess you know that and its not crucial to the proof. – mick Sep 4 '12 at 19:28
Masterful, +1 . – Jonathan Sep 4 '12 at 19:33
@mick Sorry for being sketchy. Large $u$ behavior of $\int_0^\infty \frac{u^{t-1}}{\Gamma(t)} \mathrm{d} t$ can also be obtained using Laplace's method. I used Euler-Maclaurin formula. Could you please tell me more precisely which steps you did not get. Those at the end of the post, or some others? – Sasha Sep 4 '12 at 19:37
$$ \int_0^\infty \left( \sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)!} \frac{1}{(2n)^t} \right) \mathrm{d} t = \int_0^\infty \left( \sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)!} \frac{1}{\Gamma(t)} \int_0^\infty u^{t-1} \mathrm{e}^{-2n u} \mathrm{d} u \right) \mathrm{d} t = \int_0^\infty \int_0^\infty \frac{u^{t-1}}{\Gamma(t)} \left( \sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)!} \mathrm{e}^{-2n u} \right) \mathrm{d} u \mathrm{d} t $$ The latter sum evaluates to $\frac{\cos(x \mathrm{e}^{-u})-1}{x}$. – Sasha Sep 4 '12 at 20:00
@Mick This trick of summation by integral representation can be (partially) automated, esp. using multimensional residues - see my post here. – Bill Dubuque Sep 5 '12 at 17:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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