TITLE: approximating the sum of an alternating series
QUESTION [0 upvotes]: I needed to prove that the series $$\dfrac{(-1)^{(n+1)}\ln(n)}{n}$$
starting from $n=2$, converges  and then find the approximation of the sum of the series. I managed to prove the convergence using the Leibniz alternating series test.
But now I'm kind of stuck with the next question, how do i find the approximation of the sum?
I do know that the sum resides in   $a_2< S <0$ using the Leibniz alternating series test. But is that considered an approximation?

REPLY [1 votes]: An excellent approximation to the sum of the series as $n\to\infty$ is
$$
\frac12 (\ln 2)^2 - \gamma \approx -0.1599
$$ 
where $\gamma$ is the Euler constant $$
\gamma = \lim_{n\to\infty} \left(\sum_{k=1}^n \frac1k -\ln(n)\right)$$
Possibly your problem poser meant to ask for a good asymptotic form for the value of the $n$-th sum in the series, that is, an approximation, in terms of $n$, of 
$$
S_n = \sum_{k=2}^n (-1)^{1+k}\frac{\ln(k)}{k}
$$
That is quite a meaty problem, but might be doable without very advanced techniques.
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