How to prove this ? $$-\frac\pi2 = \lim_{x\to\infty}\sum_{n=1}^{\infty}(-1)^n \frac{x^{2n-1}}{(2n)! \ln 2n}$$
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1$\begingroup$ I tried to typeset your equation more nicely. Please check if I did right. $\endgroup$– martiniSep 4, 2012 at 16:14
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8$\begingroup$ What is the source? $\endgroup$– ThéophileSep 4, 2012 at 16:26
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1$\begingroup$ 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.) $\endgroup$– Thomas AndrewsSep 4, 2012 at 16:30
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5$\begingroup$ @ThomasAndrews: it looks rather straightforward to guess which infinity mick has in mind... $\endgroup$– FabianSep 4, 2012 at 16:34
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5$\begingroup$ 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: $\,x\to -\infty\,$ $\endgroup$– DonAntonioSep 4, 2012 at 16:37
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1 Answer
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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: $$ \begin{eqnarray} \lim_{x \to \infty} \sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)! \cdot \ln(2n)} &=& \lim_{x \to \infty} \int_0^\infty \mathrm{e}^{u} \frac{\cos\left(x \mathrm{e}^{-u}\right)-1}{x} \mathrm{d} u = \lim_{x \to \infty} \int_1^\infty \frac{\cos\left(x/w\right)-1}{x} \mathrm{d} w \\ &=& \lim_{x \to \infty} \int_{1/x}^\infty \left(\cos\left(\frac{1}{v}\right)-1\right) \mathrm{d} v = \int_{0}^\infty \left(\cos\left(\frac{1}{v}\right)-1\right) \mathrm{d} v \\ &=& -\frac{\pi}{2} \end{eqnarray} $$
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1$\begingroup$ 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. $\endgroup$– mickSep 4, 2012 at 19:28
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$\begingroup$ @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? $\endgroup$– SashaSep 4, 2012 at 19:37
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4$\begingroup$ $$ \int_0^\infty \left( \sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)!} \frac{1}{(2n)^t} \right) \mathrm{d} t = \int_0^\infty \left( \sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)!} \frac{1}{\Gamma(t)} \int_0^\infty u^{t-1} \mathrm{e}^{-2n u} \mathrm{d} u \right) \mathrm{d} t = \int_0^\infty \int_0^\infty \frac{u^{t-1}}{\Gamma(t)} \left( \sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)!} \mathrm{e}^{-2n u} \right) \mathrm{d} u \mathrm{d} t $$ The latter sum evaluates to $\frac{\cos(x \mathrm{e}^{-u})-1}{x}$. $\endgroup$– SashaSep 4, 2012 at 20:00
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4$\begingroup$ @Mick This trick of summation by integral representation can be (partially) automated, esp. using multimensional residues - see my post here. $\endgroup$ Sep 5, 2012 at 17:14