Determine $\sum^{\infty}_{n=1} \frac{(-1)^{n-1}}{\ln(n+4)}$ Determine $$\sum^{\infty}_{n=1} \frac{(-1)^{n-1}}{\ln(n+4)}$$
I want to use the alternating series test, and I have already computed the limit as $n\to\infty = 0$. Now I have to show the the series is decreasing. How can I do this? 
In an example I found online, they attempted to show that the series $\sum^{\infty}_{n=1} \dfrac{(-1)^{n+1}\ln(n)}{n}$ is decreasing, and how they did that was by taking the derivative of $\dfrac{\ln(n)}{n}$, and showing that it is negative. Is this correct. Why did they leave out the $(-1)^{n+1}$?
 A: Yes, what they did is correct. You leave the $-1$ out because you know exactly how that term behaves... you need to know how the rest of the term behaves without the annoyance of an oscillating term.
A: Read first the comment of @Steven Stadnicki. In addition, a series cannot be decreasing (since if it exists, it is a real number). Instead, you should address the term decreasing to the sequence $(a_n)$ where $a_n=\frac{1}{\ln(n+4)}$. 
If we write $a_n=\frac{1}{\ln(n+4)}$ then $a_{n+1}=\frac{1}{\ln(n+5)}$. Since $n+4<n+5$ and $\ln$ is increasing, we get $0<\ln(n+4)<\ln(n+5)$ and so 
$$a_n=\frac{1}{\ln(n+4)}>\frac{1}{\ln(n+5)}=a_{n+1}\qquad \forall n\in\Bbb N$$
showing that the sequence $(a_n)$ is decreasing.
NOTE: I am confused with the example you presented as it is different with the one you posted. One thing more, its nonsense to talk about the derivative with respect to $n$.
A: The given series is convergent by Dirichlet's test, since $\{(-1)^n\}_{n\geq 1}$ is a sequence with bounded partial sums and $\frac{1}{\log(n+4)}$ is decreasing towards zero. About its exact value (this is my personal interpretation of determine) we may notice that
$$ \frac{1}{\log(n)} = \int_{1}^{+\infty}n^{1-s}\,ds $$
hence:
$$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{\log(n+4)}=\frac{3}{2\log 2}-\frac{1}{\log 3}+\int_{1}^{+\infty}\left[-1+(1-2^{2-s})\zeta(s-1)\right]\,ds $$
that can be approximated with
$$ \frac{3}{2\log 2}-\frac{1}{\log 3}-\frac{1}{2}\int_{1}^{+\infty}\exp\left((1-s)\log\frac{\pi}{2}\right)\,ds =\frac{3}{\log 4}-\frac{1}{\log 3}-\frac{1}{2\log\frac{\pi}{2}}.$$
