How to evaluate $\sum_{n=1}^{\infty}a_n$? If 
$$a_{n}=1-\frac{1}{2}+\frac{1}{3}-\cdots +\frac{\left ( -1 \right )^{n-1}}{n}-\ln 2$$
then how to eveluate 
$$\sum_{n=1}^{\infty}a_n$$
does it converge?
 A: Note that we have $\displaystyle \int_0^1 t^{k-1}=\frac{1}{k}$ for $k\geq 1$. Hence 
$$a_n=\int_0^1 \sum_{k=1}^{n}(-t)^{k-1}dt-\int_0^{1}\frac{dt}{1+t}$$ 
We compute easily that $\displaystyle a_n=-\int_0^{1}\frac{(-t)^n}{1+t}dt$. 
Hence
$$\sum_{m=1}^n a_m=-\int_0^1(\sum_{m=1}^n (-t)^m)\frac{dt}{1+t}=\int_0^1\frac{tdt}{(1+t)^2}+(-1)^{n}\int_0^1\frac{t^{n+1}}{(1+t)^2}dt=I+b_n$$
We have $|b_n|\leq \frac{1}{n+2}$, hence $b_n\to 0$; I leave to you the computation of $I$. 
A: By using Taylor expansion, we have
\begin{align}
a_{n}=(-1)^{n+1}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)
\end{align}
which alternates signs and converges to zero (note that $|a_{n}|<\frac{1}{n+1}$.) Also, 
\begin{align}
|a_{n}|&=\left(\frac{1}{n+1}-\frac{1}{n+2}\right)+\left(\frac{1}{n+3}-\frac{1}{n+4}\right)+\cdots \\
&>\left(\frac{1}{n+2}-\frac{1}{n+3}\right)+\left(\frac{1}{n+4}-\frac{1}{n+5}\right)+\cdots=|a_{n+1}|
\end{align}
so the series converges by alternating series test. 
But it seems to be difficult to find closed form.
A: Let's first consider the series $\sum_{n=0}^{\infty} b_n,$ where $b_n = \frac{(-1)^{n+1}}{n}$ Obviously this is an alternating sequence and converges to $\ln 2$. Note that $(a_n + \ln 2)$ is the partial sum of the above-mentioned series. And we have:
$$ \big|(a_n + \ln 2) - \ln(2)\big| \le \frac{1}{n+1} \implies \big | a_n \big| \le \frac{1}{n+1}$$
Now using this it's fairly easy to conclude that if $a_{n+1}$ has an opposite sign of $a_n$. Additional this proves that $\lim_{n \to \infty} \big | a_n \big | = 0$
Similarly we have that: 
$$\big | (a_n+\ln 2) - \ln 2) \big | > \big | (a_{n+1} + \ln 2) - \ln 2) \big |\implies \big | a_n \big | > \big | a_{n+1} \big |$$
Therefore as the sequence is alternating, decreasing and it's converging to $0$, by Alternating Series Test we have that it converges.
