The sum of $(-1)^n \frac{\ln n}{n}$ 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
$$\sum_2^{\infty}(-1)^n \frac{ \ln n}{n}=\gamma\ln 2 -\frac{ \ln^2 2}{2}$$

Consider
$$P=\frac{\ln 2}{2}+\frac{\ln 3}{3}+\frac{\ln 4}{4}+...+\frac{\ln 2n}{2n}$$
$$Q=\frac{2\ln 2}{2}+\frac{2\ln 4}{4}+\frac{2\ln 6}{6}+...+\frac{2\ln 2n}{2n}=\left( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n} \right)\ln 2+\frac{\ln 2}{2}+\frac{\ln 3}{3}+\frac{\ln 4}{4}+...+\frac{\ln n}{n}$$
$$Q-P=\sum\limits_{j=2}^{2n}{\frac{\left( -1 \right)^{j}}{j}\ln j}=\left( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n} \right)\ln 2-\left( \frac{\ln \left( n+1 \right)}{n+1}+\frac{\ln \left( n+2 \right)}{n+2}+\frac{\ln \left( n+3 \right)}{n+3}+...+\frac{\ln 2n}{2n} \right)$$
$$=\left( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n \right)\ln 2+\ln 2\ln n-\sum\limits_{j=1}^{n}{\frac{\ln n}{n+j}}-\sum\limits_{j=1}^{n}{\frac{1}{n}\cdot \frac{\ln \left( 1+\frac{j}{n} \right)}{1+\frac{j}{n}}}$$
$$\sum\limits_{j=2}^{+\infty }{\frac{\left( -1 \right)^{j}}{j}\ln j}=\gamma \ln 2-\int\limits_{0}^{1}{\frac{\ln \left( 1+x \right)}{1+x}dx}=\gamma \ln 2-\frac{\ln ^{2}2}{2}$$
Done
