<s>[INST] 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?
 [/INST] Write, as usual,
$$H_n=\sum_{k=1}^n\frac1 k.$$
Then
$$\frac11+\frac13+\frac15+\cdots+\frac1{2n-1}=H_{2n}-\frac{H_n}2.$$
In your notation,
$$S_{3n}=H_{4n}-\frac{H_{2n}}{2}-\frac{H_n}{2}.$$
But
$$H_n=\ln n+\gamma+O(1/n)$$
where $\gamma$ is Euler's constant, so
$$S_{3n}=\ln 4n-\frac{\ln 2n}{2}-\frac{\ln n}{2}+O(1/n)=\frac{3\ln2}2
+O(1/n)$$
so the limit is $\frac32\ln2$.
</s>