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I'm stuck trying to show that $$\sum_{n=2}^{\infty} (-1)^n \frac{\ln n}{n}=\gamma \ln 2- \frac{1}{2}(\ln 2)^2$$

This is a problem in Calculus by Simmons. It's in the end of chapter review and it's associated with the section about the alternating series test. There's a hint: refer to an equation from a previous section on the integral test. Specifically:

$$L=\lim_{n\to\infty} F(n)=\lim_{n\to\infty}\left[a_1+a_2+\cdots+a_n-\int_1^n\! f(x)\,\mathrm{d}x\right]$$

Here, $\{a_n\}$ is a decreasing sequence of positive numbers and $f(x)$ is a decreasing function such that $f(n)=a_n$, and $\gamma$ is this limit in the case that $a_n=\frac{ 1}{n}$.

New users can't answer their own questions inside of 8 hours, so I'm editing my question to reflect the answer.

Ok, I got it.

Following the hint in the book

$$L=\lim_{n\to\infty}\left[\frac{\ln 2}{2}+\frac{\ln 3}{3}+\cdots+\frac{\ln n}{n}-\int_2^n\! \frac{\ln x}{x}\,\mathrm{d}x\right]$$

$$=\lim\left[\frac{ \ln 2}{2}+\cdots+\frac{ \ln n}{n}-\left.\frac{ \ln^2x}{2}\right|_2^n\right]$$

The partial sum for the positive series is: $$\left(\frac{\ln^2n}{2}-\frac{\ln^2}{2}\right)+L+o(1)$$

Returning to the original, alternating series: $$-S_{2n}=\frac{ -\ln 2}{2}+\frac{ \ln 3}{3}-\frac{\ln 4}{4}+\frac{\ln 5}{5}-\cdots$$ $$=\frac{\ln 2}{2}+\frac{\ln 3}{3}+\cdots+\frac{\ln 2n}{2n}-2\left(\frac{\ln 2}{2}+\frac{\ln 4}{4}+\cdots+\frac{\ln 2n}{2n}\right)$$

Consider the partial sum in parentheses $$ \frac{\ln 2}{2}+\frac{\ln 4}{4}+\cdots+\frac{\ln 2n}{2n}=\frac{\ln 2}{2}+\frac{\ln 2 +\ln 2}{4}+\frac{\ln 2+\ln 3}{6}+\cdots+\frac{\ln 2+\ln n}{2n}$$ $$=\frac{1}{2}\left(\ln 2\left(1+\frac{ 1}{2}+\cdots+\frac{ 1}{n}\right)+\left(\frac{ \ln 2}{2}+\frac{ \ln 3}{3}+\cdots+\frac{ \ln n}{n}\right)\right)$$

Now, plug that back in $$-S_{2n}=\left(\frac{\ln 2}{2}+\cdots+\frac{ \ln 2n}{2n}\right)-\ln 2\left(1+\frac{ 1}{2}+\cdots+\frac{ 1}{n}\right)-\left(\frac{ \ln 2}{2}+\frac{ \ln 3}{3}+\cdots+\frac{ \ln n}{n}\right)$$ $$=\frac{ \ln^2(2n)}{2}-\frac{ \ln^2 2}{2}+L+o(1)-\ln 2\left(\ln n +\gamma+o(1)\right)-\left(\frac{ \ln^2 n}{2}-\frac{ \ln^2 2}{2}+L+o(1)\right)$$ $$=\frac{ (\ln 2 +\ln n)^2}{2}-(\ln 2)(\ln n)-\gamma\ln 2-\frac{ \ln^2 n}{2}+o(1)$$ $$=\frac{ \ln^2 2}{2}+(\ln 2)(\ln n)+\frac{ \ln^2 n}{2}-(\ln 2)(\ln n)-\gamma \ln 2 - \frac{ \ln^2 n}{2}+o(1)$$ $$-S_{2n}\to\frac{ \ln^2}{2}-\gamma\ln 2$$ Which gives the desired result $$\sum_2^{\infty}(-1)^n \frac{ \ln n}{n}=\gamma\ln 2 -\frac{ \ln^2 2}{2}$$

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I don't understand why we can start from the Limit "L",where did the formula $\displaystyle L=\lim_{n\to\infty} F(n)=\lim_{n\to\infty}\left[a_1+a_2+\cdots+a_n-\int_1^n\! f(x)\,\mathrm{d}x\right]$ come from.And why $f(n)$ must be $a_{n}$ – EvilNebula. Jan 26 at 17:03

Let's evaluate this more generally. What is $$\sum_{n=1}^\infty \frac{(-1)^n \log^k (n)}{n}$$ for integers $k$? Recall the Dirichlet eta function $$\eta(s)=\sum_{n=1}^\infty \frac{(-1)^n}{n^s}=\left(1-2^{1-s}\right)\zeta(s).$$
Lets look at the expansion around $s=1$.
We have that

$$1-2^{1-s}=1-e^{-(s-1)\log2}=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\left(\log2\right)^{n}}{n!}(s-1)^{n},$$ and $$\zeta(s)=\frac{1}{s-1}+\sum_{n=0}^{\infty}(-1)^{n}\gamma_{n}\frac{(s-1)^{n}}{n!}.$$ The $\gamma_i$ are the Stieltjes Constants which we expect should come up in this problem since $$\gamma_m:=\lim_{r\to\infty} \sum_{k=1}^r \left(\frac{\log^m k}{k}-\frac{\log^m r}{(m+1)}\right).$$ Upon multiplying the two expansions we get $$\left(1-2^{1-s}\right)\zeta(s)=\sum_{n=0}^{\infty}\left(\frac{(-1)^{n}\left(\log2\right)^{n+1}}{(n+1)!}+(-1)^{n-1}\sum_{k=0}^{n-1}\gamma_{k}\frac{\log^{n-k}2}{k!(n-k)!}\right)(s-1)^{n}.$$ Notice that $$\sum_{n=1}^{\infty}\frac{(-1)^{n}\log^{k}n}{n}=\eta^{(k)}(1)$$ which is the $k^{th}$ derivative of the above expression evaluated at $1$. Consequently, it is the $k^{th}$ coefficient above multiplied by $k!$. That is we have the closed form $$\sum_{n=1}^{\infty}\frac{(-1)^{n}\log^{k}n}{n}=\frac{(-1)^{n}\left(\log2\right)^{k+1}}{k+1}+(-1)^{k-1}\sum_{j=0}^{k-1}\gamma_{j}\binom{k}{j}\log^{k-j}2.$$

Hope that helps,

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Here is an another way to show the identity in calculus level, although not as simple as the solution from the hint. A brutal force works here. Let $$H_n = \sum_{k=1}^{n} \frac{1}{k}$$ be the $n$-th harmonic number and $$ A_{r,n} = \sum_{k=1}^{n} \frac{1}{n} \frac{\left( \log ( 1 + \frac{k}{n} ) \right)^{r}}{1 + \frac{k}{n}}.$$ Then we have $$ \sum_{k=1}^{2n} \frac{(-1)^{k} \log k}{k} = (H_n - \log n) \log 2 + (\log 2 - A_{0, n}) \log n - A_{1,n}.$$ It is not hard to see that if $f$ is of class $C^1$ on $[0, 1]$, then by mean value theorem, $$\lim_{n\to\infty} n \left( \sum_{k=1}^{n} f\left( \frac{k}{n}\right) \frac{1}{n} - \int_{0}^{1} f(x) \; dx \right) = \frac{f(1) - f(0)}{2}.$$ Thus assuming the identity above, taking $n \to \infty$, we have $$ \sum_{k=1}^{\infty} \frac{(-1)^{k} \log k}{k} = \gamma \log 2 - \frac{1}{2} (\log 2)^{2}.$$ So it remains to show the identity. The key observation that leads to this result is the identity $$ \sum_{k=1}^{2n} \frac{(-1)^{k} \log k}{k} = \sum_{k=1}^{n} \frac{\log (2k)}{k} - \sum_{k=1}^{2n} \frac{\log k}{k},$$ which is obtained by splitting even terms and odd terms.

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+1. Also, just thought I would also add that your claim that uses the mean value theorem is the first-order approximation from the Euler-Maclaurin summation formula. – Mike Spivey May 24 '11 at 13:40
I'm very interested in the generalization of your fourth equation to any interval $[a,b]$. What would it be? – Pedro Tamaroff Aug 1 '12 at 0:18

Hint: Consider $ \frac{\ln(2k)}{2k} - \frac{\ln(2k+1)}{2k+1}$.

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For more literate solution: Consider $$\kappa(s)=\sum_{n=2}^{\infty} \frac{(-1)^{n+1}\ln n}{n^s}$$

when $s=1$ we can use abel's theorem to verified the convergence since the above series is absolutely convergent for $s\geq 1$ Now suppose that $s>1$. Define $$\psi(s)=\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n^s}$$ The absolute convergence justified the integration of the series $\kappa(s)$, so that we have $\psi'(s)= \kappa(s)$.

And so we wish to estimate $\psi'(1+)$. Notice also that $\phi(s)=(1-2^{1-s})\zeta(s)$, where $\zeta(s)$ is Riemann Zeta Function. We use the following result $$\gamma = \lim_{s\rightarrow 1+} \zeta(s) - \frac{1}{s-1}$$ or we can say $$\zeta(s)=\gamma+\left(\frac{1}{s-1}\right)+O({s-1})$$ For proof see: J. Sondow, An antisymmetric formula for Euler's constant, Mathematics Magazine, (1998), vol. 71, number 3, pp. 219-220.

Consider the expansion of

$$1- 2^{1-s}=1-e^{\ln 2^{1-s}}=(s-1) \ln 2 - \frac{(s-1)^2 \ln ^2 2 }{2} + \cdots$$

and for $s\rightarrow 1^+$ we can write $\displaystyle 1-2^{1-s}=(s-1) \ln 2 - \frac{(s-1)^2 \ln ^2 2 }{2}+O(s^3)$. So that

$$\psi(s)=(1-2^{1-s})\zeta(s)=\left((s-1) \ln 2 - \frac{(s-1)^2 \ln ^2 2 }{2}+ O(s^3)\right)\left(\gamma+\left(\frac{1}{s-1}\right)+O({s-1})\right)$$

finishing the expansion, differentiate and taking $s\rightarrow 1^+$ we have the result.

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Utilizing Dirichlet eta function is also a standard story, but it seems far beyond the scope of basic calculus. – Sangchul Lee May 24 '11 at 6:49
yes, i agree :) , for -inside the scope- your proof is really recommended. – Ajat Adriansyah May 24 '11 at 8:17
There are a few problems here: (1) Writing $$\zeta(s)=\gamma +O\left(\frac{1}{s-1}\right)$$ makes little sense, and is not going to give the desired solution. Instead, what the limit gives us is $$\zeta(s)=\frac{1}{s-1}+\gamma+O(s-1).$$ You need this expansion to evaluate things later on (the part you skipped). (2) Why $\psi'(1+)$? You don't need the $+$, its just $\psi^'(1)$ in your notation since the function $\psi(s)$ is analytic in the RHP. Same thing with at the end, you can just take $s\rightarrow 1$, no need to come from the right. (Also, the conventional notation is $\eta(s)$) – Eric Naslund May 25 '11 at 16:00
Also, I am voting -1 mainly because of the first line, "for a more literate solution...," which is then followed by many errors. Here is another: "abel's theorem to verified the convergence.." No. You use the alternating series test, or Dirichlets test to verify convergence. Abel's limit theorem is also only for power series and wouldn't apply here anyway. It tells us that we can switch the order of the limit and the summation if the series converges, but again that is for power series. Please clean things up so I can upvote your answer instead. – Eric Naslund May 25 '11 at 16:12
Dear Eric Naslund, I'm sorry i wasn't around for few days.. I agree about the big O notation, I fixed it.. also I was using Abel Summation By part first before applying the test.. – Ajat Adriansyah Jun 2 '11 at 1:05

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