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I was trying to solve this problem: Closed form for $\int_0^1\log\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\mathrm dx$

In the procedure I followed, I came across the following sum: $$\sum_{k=1}^{\infty} (-1)^{k-1}k\left(\frac{\ln(2k+1)}{2k+1}-\frac{\ln(2k-1)}{2k-1}\right)$$

I cannot think of any approaches which would help me in evaluating the sum.

Any help is appreciated. Thanks!

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  • $\begingroup$ I must confess that, taking into account the answers you received, I have a problem : the numerical evaluation of the sum shows a convergence to a value of $0.391594$. $\endgroup$ – Claude Leibovici Jul 17 '14 at 9:56
  • $\begingroup$ @ClaudeLeibovici Looks like we have a strongly oscillating sum here, I also agree with your observation. $\endgroup$ – AlexR Jul 17 '14 at 10:14
  • $\begingroup$ @AlexR. Yes, this is what happens ! Thanks for supporting the old man ! Cheers :) $\endgroup$ – Claude Leibovici Jul 17 '14 at 10:19
  • $\begingroup$ @ClaudeLeibovici: Thanks for chipping in, I was about to lose hope in the procedure I followed for the linked problem. :) $\endgroup$ – Pranav Arora Jul 17 '14 at 10:29
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    $\begingroup$ The series is convergent using the alternating series test:$$k\left(\frac{\ln(2k+1)}{2k+1}-\frac{\ln(2k-1)}{2k-1}\right) \sim \frac{\ln k}{2k}$$ as $k$ tends to $+\infty$. $\endgroup$ – Olivier Oloa Jul 17 '14 at 10:44
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The terms of the series vanish so we will look only at the partial sums where $n$ is odd. $$ \begin{align} &\sum_{k=1}^\infty(-1)^{k-1}k\left(\frac{\log(2k+1)}{2k+1}-\frac{\log(2k-1)}{2k-1}\right)\\ &=\lim_{n\to\infty}\left(\sum_{k=1}^n(-1)^{k-1}k\frac{\log(2k+1)}{2k+1}-\sum_{k=0}^{n-1}(-1)^k(k+1)\frac{\log(2k+1)}{2k+1}\right)\tag{1}\\ &=\lim_{n\to\infty}\left((-1)^{n-1}n\frac{\log(2n+1)}{2n+1}+\sum_{k=1}^{n-1}(-1)^{k-1}\log(2k+1)\right)\tag{2}\\ &=\lim_{n\to\infty}\left((2n+1)\frac{\log(4n+3)}{4n+3}+\sum_{k=1}^{n}\log\left(\frac{4k-1}{4k+1}\right)\right)\tag{3}\\ &=\lim_{n\to\infty}\log\left((4n+3)^{\frac{2n+1}{4n+3}}\cdot\prod_{k=1}^{n}\frac{k-\frac14}{k+\frac14}\right)\tag{4}\\ &=\lim_{n\to\infty}\log\left(2\sqrt{n}\frac{\Gamma(n+\frac34)}{\Gamma(\frac34)}\frac{\Gamma(\frac54)}{\Gamma(n+\frac54)}\right)\tag{5}\\ &=\log\left(2\frac{\Gamma(\frac54)}{\Gamma(\frac34)}\right)\tag{6}\\ &=\log\left(\frac12\frac{\Gamma(\frac14)}{\Gamma(\frac34)}\right)\tag{7}\\ \end{align} $$ Explanation:
$(1)$: write series as the limit of the partial sums, split into two sums, reindex the second sum
$(2)$: recombine the two sums
$(3)$: substitute $n\mapsto2n+1$ and combine pairs of terms of the sum
$(4)$: write a sum of logs as a log of a product
$(5)$: $\lim\limits_{n\to\infty}\frac1{2\sqrt{n}}(4n+3)^{\frac{2n+1}{4n+3}}=1$, write the product as a ratio of Gamma functions
$(6)$: Gautschi's inequality
$(7)$: $x\Gamma(x)=\Gamma(x+1)$

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As said in comments, the sum oscillates between two values and converges to the value $0.391594$.

If we expand in the same spirit as Larsen, what we can see is that $$F(n)=\sum_{k=1}^{n} (-1)^{k-1}k\left(\frac{\ln(2k+1)}{2k+1}-\frac{\ln(2k-1)}{2k-1}\right)$$ can be written as $$F(n)=\sum_{k=1}^{n-1} (-1)^{k-1}\log(2k+1)-\frac{(-1)^n n \log(2n+1)}{2n+1}$$

Added after AlexR's answer and comments

Let us consider the last term in AlexR' answer $$\Delta = (m+1) \left(\frac1{2m+3} \ln(2m+3) - \frac1{2m+1} \ln(2m+1)\right) $$ and let us expand it as a Taylor series for large values of $m$. We so obtain $$\Delta =\frac{\log \left(\frac{1}{m}\right)+1-\log (2)}{2 m}+\frac{-\log \left(\frac{1}{m}\right)-2+\log (2)}{2 m^2}+O\left(\left(\frac{1}{m}\right)^3\right)$$

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  • $\begingroup$ Found a nice estimate-based approach to circumvent Taylor ;) Maybe Taylor will be useful in finding the limit (Think of $S_m$ expanded at $m = \infty$)... (+1) $\endgroup$ – AlexR Jul 17 '14 at 10:59
  • $\begingroup$ Be sure it is for me a real pleasure to work this problem at the same time as you. Sinergy has been proven (if required). Thank you. $\endgroup$ – Claude Leibovici Jul 17 '14 at 11:05
  • $\begingroup$ Happy to work with you as well. I proceeded to find the limit using my partial sums: W|A finally found something $$\lim_{m\to\infty} S_m = \ln\left(\frac{\Gamma(\frac14)}{2\Gamma(\frac34)}\right)$$ :-) $\endgroup$ – AlexR Jul 17 '14 at 11:08
  • $\begingroup$ @AlexR. This is just great ! Congratulations ! $\endgroup$ – Claude Leibovici Jul 17 '14 at 11:34
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On the limit
Experimenting a bit I find $$\lim_{m\to\infty} S_m = \ln\left(\frac{\Gamma(\frac14)}{2\Gamma(\frac34)}\right) = 0.391594392706836...$$ To be the exact limit

On convergence
Let $a_k := \frac{\ln(2k-1)}{2k-1}$ then $$\begin{align*} S_m & = \sum_{k=1}^m (-1)^{k-1} k (a_{k+1} - a_k) \\ & = \sum_{k=2}^{m+1} (-1)^k (k-1) a_k + \sum_{k=1}^m (-1)^k k a_k \\ & \stackrel{a_1 = 0}= \sum_{k=2}^m (-1)^k ((k-1) a_k + k a_k) + (-1)^{m+1} ma_{m+1} \\ & = \sum_{k=2}^m (-1)^k \ln(2k-1) + (-1)^{m+1} \frac{m}{2m+1} \ln(2m+1) \end{align*}$$ The problem all answerers overlooked is the factor of the last summand not being $\frac1{2m+1}$ but $\frac{m}{2m+1}$.


Let's first try to prove that the sequence is cauchy: $$\begin{align*} |S_{m+1}-S_m| & = |(-1)^{m+1} \ln(2m+1) \\ &\left. \qquad + (-1)^m \frac{m+1}{2m+3} \ln(2m+3) - (-1)^{m+1} \frac m{2m+1} \ln(2m+1) \right| \\ & = \left| (m+1) \left(\frac1{2m+3} \ln(2m+3) - \frac1{2m+1} \ln(2m+1)\right) \right| \\ & \le (m+1) \ln(2m+3) \left(\frac1{2m+1} - \frac1{2m+3}\right) \\ & = \ln(2m+3) (m+1) \frac2{4m(m+2) + 3} \\ & \le \ln(2m+3) \frac1{2m} \to 0 & (m\to\infty) \end{align*}$$

This proves convergence. The actual value seems a tad bit harder to prove.

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