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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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I tried to typeset your equation more nicely. Please check if I did right. –  martini Sep 4 '12 at 16:14
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What is the source? –  Théophile Sep 4 '12 at 16:26
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@ThomasAndrews: it looks rather straightforward to guess which infinity mick has in mind... –  Fabian Sep 4 '12 at 16:34
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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
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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
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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: $$ \begin{eqnarray} \lim_{x \to \infty} \sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)! \cdot \ln(2n)} &=& \lim_{x \to \infty} \int_0^\infty \mathrm{e}^{u} \frac{\cos\left(x \mathrm{e}^{-u}\right)-1}{x} \mathrm{d} u = \lim_{x \to \infty} \int_1^\infty \frac{\cos\left(x/w\right)-1}{x} \mathrm{d} w \\ &=& \lim_{x \to \infty} \int_{1/x}^\infty \left(\cos\left(\frac{1}{v}\right)-1\right) \mathrm{d} v = \int_{0}^\infty \left(\cos\left(\frac{1}{v}\right)-1\right) \mathrm{d} v \\ &=& -\frac{\pi}{2} \end{eqnarray} $$

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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
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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
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$$ \int_0^\infty \left( \sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)!} \frac{1}{(2n)^t} \right) \mathrm{d} t = \int_0^\infty \left( \sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)!} \frac{1}{\Gamma(t)} \int_0^\infty u^{t-1} \mathrm{e}^{-2n u} \mathrm{d} u \right) \mathrm{d} t = \int_0^\infty \int_0^\infty \frac{u^{t-1}}{\Gamma(t)} \left( \sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)!} \mathrm{e}^{-2n u} \right) \mathrm{d} u \mathrm{d} t $$ The latter sum evaluates to $\frac{\cos(x \mathrm{e}^{-u})-1}{x}$. –  Sasha Sep 4 '12 at 20:00
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@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
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