# A curious limit for $-\frac{\pi}{2}$

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

• 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
• You have to pick an infinity, I think: $x\to +\infty$ or $x\to -\infty$. Might seem pedantic, but $x\to\infty$ means something specific. (We often write $n\to\infty$, but there, $n$ is usually a natural number, and there is only "one" infinity it can go to.) – Thomas Andrews Sep 4 '12 at 16:30
• @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

## 1 Answer

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

• 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