The sum of $(-1)^n \frac{\ln n}{n}$

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 $$\gamma_m:=\lim_{r\to\infty} \left(\sum_{k=1}^r \frac{\log^m k}{k}-\frac{\log^{m+1} r}{(m+1)}\right).$$ 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,

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.

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