How to prove this ? $$-\frac\pi2 = \lim_{x\to\infty}\sum_{n=1}^{\infty}(-1)^n \frac{x^{2n-1}}{(2n)! \ln 2n}$$
A curious limit for $-\frac{\pi}{2}$
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1I tried to typeset your equation more nicely. Please check if I did right. – 2012-09-04
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0Personally, I'd put the $(-1)^n$ in the numerator. – 2012-09-04
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8What is the source? – 2012-09-04
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1You 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.) – 2012-09-04
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5@ThomasAndrews: it looks rather straightforward to guess which infinity mick has in mind... – 2012-09-04
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5Unless 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: $\,x\to -\infty\,$ – 2012-09-04
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0@Fabian Yes, as I said, it may seem pedantic, but it is important to get used to making clear distinctions when talking about limits. Not knowing OP, but based on the original posting formatting, I assumed he was inexperienced, and therefore would be helped by reminding him of the distinction. – 2012-09-04
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0The source is me. In the original i typed +oo. I believe -oo would give + $\pi/2$ btw. ( its been a long long time since i came up with this lim ) Thanks for the Tex edit. I prefer the (-1) above too. – 2012-09-04
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0lol its trivial that -oo gives the same result times -1. :) – 2012-09-04
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0@mick Do you remember in what context you came up with this limit? This could help us figure out what tools could be relevant in finding the limit. – 2012-09-04
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1As 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 ). – 2012-09-04
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0Numerically [Wolfram](http://www.wolframalpha.com/input/?i=%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%28-1%29%5En+%5Cfrac%7B%2810%29%5E%7B2n-1%7D%7D%7B%282n%29%21+%5Cln+2n%7D) seems to confirm that the limit is indeed $\frac{\pi}{2}$. The question is how to deal with the logarithm on the bottom. – 2012-09-04
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0Your question is the same as trying to show that $$\int_{0}^{\infty}\sum_{n=1}^{\infty}(-1)^{n}\frac{x^{2n}}{(2n)\left(2n\right)!\log(2n)}dx=\frac{-\pi}{2} $$ I am not sure how to do this, but I have seen similar integrals such as $$\int_{0}^{\infty}\sum_{n=1}^{\infty}(-1)^{n}\frac{x^{2n-2}}{(2n-1)!}dx=-\frac{\pi}{2} $$ which follows from basic complex analysis. Could you elaborate more on where this integral came from? – 2012-09-04
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0Surprisingly the formula is also in http://math.eretrandre.org/tetrationforum/showthread.php?tid=743 dated 7.Aug.2012 but was not discussed there. – 2012-10-04
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0Not a surprise. I asked permission to post it here. Do I need to give credit to others in my questions ? – 2012-10-04
1 Answers
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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1Euh 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. – 2012-09-04
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2Masterful, +1 . – 2012-09-04
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0@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](http://en.wikipedia.org/wiki/Laplace%27s_method). I used [Euler-Maclaurin formula](http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula). Could you please tell me more precisely which steps you did not get. Those at the end of the post, or some others? – 2012-09-04
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0@Jonathan Thanks for the upvote and for your kind words. – 2012-09-04
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0@ Sahsha I feel silly for not understanding , but you already lost me at the double integral. blush – 2012-09-04
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4$$ \int_0^\infty \left( \sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)!} \frac{1}{(2n)^t} \right) \mathrm{d} t = \int_0^\infty \left( \sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)!} \frac{1}{\Gamma(t)} \int_0^\infty u^{t-1} \mathrm{e}^{-2n u} \mathrm{d} u \right) \mathrm{d} t = \int_0^\infty \int_0^\infty \frac{u^{t-1}}{\Gamma(t)} \left( \sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)!} \mathrm{e}^{-2n u} \right) \mathrm{d} u \mathrm{d} t $$ The latter sum evaluates to $\frac{\cos(x \mathrm{e}^{-u})-1}{x}$. – 2012-09-04
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0Thanks. Off course. Excellent use of the gamma function. – 2012-09-04
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4@Mick This trick of summation by integral representation can be (partially) automated, esp. using multimensional residues - see [my post here.](http://math.stackexchange.com/a/3720/242) – 2012-09-05
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0Nice proof! (+1) – 2012-09-07
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0uh, i missed this gem (+1) – 2018-06-12