Another "can't use Alt. Series Conv. test" (How do I determine if it converges or diverges?) Stumped on yet another absolute convergence problem.
(Converge conditionally, absolutely, or diverges)
$$\sum_{n=1}^\infty \frac{(-1)^n \ln(n)}{n} $$
First, I tested for absolute convergence, and the series reduced to $\frac{\ln(n)}{n}$   .....Using direct comparison, $\frac{\ln(n)}{n} > \frac{1}{n}$ I determined this diverges.  So, it does not converge absolutely.
Next, to use the alternating series test on the original series, you need to make sure the magnitude of the terms are decreasing.  Well, they are not:
$$\sum_{n=1}^\infty \frac{(-1)^n \ln(n)}{n} = -0 + .35 - .37 + .35.... $$
So, without using the alternating series test, how do I determine if the orig. series converges or diverges?
How can I use the divergence test (Nth term test)?  I'm not sure what the nth term tends to with that (-1) oscillation.  Just focus on $\frac{\ln(n)}{n}$  But we established that diverges, right?   But, I am told this series converges conditionally.
Do I use integral test?
 A: Suppose you break the series into its first 100 terms (which has a finite sum) and the remaining infinite series. You can then use the alternating series test on that remaining series, in which the terms DO decrease (although you'll need to prove this). 
This is a general trick for infinite series: what happens in the first few terms never affects convergence, so you're welcome to apply any of the tests starting at the $k$th term (for any finite value of $k$ that you like). This can be pretty useful. 
A: The series is not absolutely convergent by comparison with the harmonic series (that is divergent), but since $(-1)^n$ has bounded partial sums and $\frac{\log n}{n}$ decreases towards zero from some point on, the series is conditionally convergent by Dirichlet's test. We also have:
$$ S=\sum_{n\geq 1}\frac{(-1)^n \log n}{n} = \left.\frac{d}{d\alpha}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^\alpha}\right|_{\alpha=1^+}=\left.\frac{d}{d\alpha}\left[\left(1-\frac{1}{2^{\alpha-1}}\right)\zeta(\alpha)\right]\right|_{\alpha=1^+}$$
that can be computed through the reflection formula and the residue of the Riemann $\zeta$ function at $\alpha=1$:

$$ S=\gamma\log(2)-\frac{\log^2(2)}{2} $$

where $\gamma$ is the Eulero-Mascheroni constant.

A more elementary approach comes from coupling consecutive terms:
$$ \frac{\log n}{n}-\frac{\log(n+1)}{n+1} = \frac{\log n}{n(n+1)}+\frac{\log\left(1+\frac{1}{n}\right)}{n+1}\leq \frac{1+\log n}{n(n+1)} $$
but $\sum_{n\geq 1}\frac{1+\log n}{n(n+1)}$ is quite trivially absolutely convergent.
A: $\sum_{n=1}^\infty \frac{(-1)^n \ln(n)}{n}
$
Another way to deal with
$\sum_{n=1}^\infty (-1)^n a_n
$
where
$a_n \to 0$
is to consider
$s_m
=\sum_{n=1}^{2m+1} (-1)^n a_n
=a_1+\sum_{n=2}^{2m+1} (-1)^n a_n
=a_1+\sum_{n=1}^{m}(a_{2n}-a_{2n+1})
$.
If $s_m$ converges,
then the overall series converges.
In this case,
$\begin{array}\\
a_{2n}-a_{2n+1}
&=\frac{\ln(2n)}{2n}-\frac{\ln(2n+1)}{2n+1}\\
&=\frac{(2n+1)\ln(2n)-2n\ln(2n+1)}{2n(2n+1)}\\
&=\frac{2n(\ln(2n)-\ln(2n+1))+\ln(2n)}{2n(2n+1)}\\
&=\frac{2n(-\ln(1+1/(2n)))+\ln(2n)}{2n(2n+1)}\\
&\gt\frac{2n(-1/(2n))+\ln(2n)}{2n(2n+1)}
\qquad\text{since }\ln(1+x) < x\\
&=\frac{-1+\ln(2n)}{2n(2n+1)}\\
\text{and}\\
a_{2n}-a_{2n-1}
&\lt\frac{\ln(2n)}{2n(2n+1)}
\qquad\text{since }\ln(1+x) > 0\\
\end{array}
$
Since the sum
of both the
lower and upper bounds converge,
the series converges.
