Compute : $1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + \cdots$ 
Compute : $1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + \cdots$

My try
It can be verified that $\lim_{k \to \infty} S_{3k} < + \infty$ and $\lim_{k \to \infty} S_{3k} = \lim_{k \to \infty} S_{3k+1} = \lim_{k \to \infty} S_{3k+2}$.
So letting $a_n := S_{3n}$, $a_{n+1} - a_n = \frac{1}{4n+1} + \frac{1}{4n+3} - \frac{1}{2n + 2} = \frac{8k+5}{(4k+1)(4k+3)(2k+2)}$.
Since  $\lim (a_{n+1} - a_n) = \lim S_n - a_1$, suffice to compute $\lim_{n\to\infty}(a_{n+1} - a_n)$.
$$
\begin{aligned}
\lim_{n \to \infty} (\frac{1}{4n+1} + \frac{1}{4n+3} - \frac{1}{2n + 2}) &=\lim_{n \to \infty} (\frac{1}{4n+1}) + \lim_{n \to \infty} (\frac{1}{4n+3}) - \lim_{n \to \infty} (\frac{1}{2n+2}) \\
&= \frac{5}{6}
\end{aligned}
$$
And $a_1 = S_3 = 5/6$, thus $\lim S_n = 5/3$. Am I right?

\begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
   && 1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9} + \cdots\\
   &=&1 + \frac{1}{3}-\frac{2}{4}+\frac{1}{5}+\frac{1}{7}-\frac{2}{8}+\frac{1}{9} + \cdots\\
   &=& \int_0^1\frac{1+x^2-2x^3}{1-x^4}dx=\int_0^1\frac{(1-x)(1+x+2x^2)}{(1-x)(1+x)(1+x^2)}dx\\
&=&
\int_0^1\left(\frac{1}{1+x}+\frac{x}{1+x^2}\right)dx\\
&=&\left(\ln(1+x)+\frac{1}{2}\ln(1+x^2)\right)\Big|_0^1=\frac{3}{2}\ln 2.
\end{eqnarray*}
