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?

It seems to me the pattern is you are adding the following triplets:  $\frac 1 {4k+1} +\frac 1 {4k+3}-\frac 1 {2^{k+1}} $.
If this converged we could rearrange the terms.  The  infinite sum of $\sum\frac {-1} {2^{k+1}}$ is $-1$ which is finite so that would mean the sum $\sum (\frac 1 {4k+1}+\frac 1 {4k+3})= \sum \frac 1 {2k+1} $ is finite.
But it's not because it is harmonic.
So the sum does not converge.
Unless the pattern is something else.
