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?
 A: 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 $\frac32\ln2$.
A: The sum in the question can be re-written as
$$
\begin{align}
\sum_{k=1}^\infty\left(\frac1{4k-3}+\frac1{4k-1}-\frac1{2k}\right)
&=\lim_{n\to\infty}\sum_{k=1}^n\left(\frac1{4k-3}+\frac1{4k-1}-\frac1{2k}\right)\\
&=\lim_{n\to\infty}\left(\sum_{k=1}^{4n}\frac1k-\sum_{k=1}^{2n}\frac1{2k}-\sum_{k=1}^n\frac1{2k}\right)\\
&=\frac12\lim_{n\to\infty}\left(\sum_{k=1}^{4n}\frac1k-\sum_{k=1}^{2n}\frac1k\right)+\frac12\lim_{n\to\infty}\left(\sum_{k=1}^{4n}\frac1k-\sum_{k=1}^n\frac1k\right)\\
&=\frac12\lim_{n\to\infty}\sum_{k=2n+1}^{4n}\frac nk\frac1n+\frac12\lim_{n\to\infty}\sum_{k=n+1}^{4n}\frac nk\frac1n\\
&=\frac12\int_2^4\frac1x\,\mathrm{d}x+\frac12\int_1^4\frac1x\,\mathrm{d}x\\[6pt]
&=\frac12\log(2)+\frac12\log(4)\\[9pt]
&=\frac32\log(2)
\end{align}
$$
A: 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.
A: \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*}
