<s>[INST] Find an approximate sum of the series $\sum_{k=0}^{\infty}\frac{1}{2^{2k+1}(2k+1)}$ $$
S=\sum_{k=0}^{\infty}\frac{1}{2^{2k+1}(2k+1)}\ \ \ (*)
$$
I have to find $S_n: |S-S_n|<\epsilon=10^{-4}$
Also, I cannot use expansion series for any functions.  
This is what I came up with so far:
$$
a_k=\frac{1}{2^{2k+1}(2k+1)}\leqslant\frac{1}{(2k+1)^2}=b_k\\
\sum_{k=0}^{\infty}b_k \text{ converges. Therefore, $(*)$ converges too.}
$$
Also, I noticed that $a_k$ is a monotonically decreasing function. But I still do not know how to apply all these calculations to the problem
 [/INST] $S
=\sum_{k=0}^{\infty}\frac{1}{2^{2k+1}(2k+1)}\ \ \ (*)
$
Let
$t_n
=\sum_{k=n}^{\infty}\frac{1}{2^{2k+1}(2k+1)}
$.
Then
$\begin{array}\\
t_n
&\lt\sum_{k=n}^{\infty}\frac{1}{2^{2k+1}(2n+1)}\\
&=\frac1{2n+1}\sum_{k=n}^{\infty}\frac{1}{2^{2k+1}}\\
&=\frac1{2n+1}\sum_{k=0}^{\infty}\frac{1}{2^{2k+2n+1}}\\
&=\frac1{(2n+1)2^{2n+1}}\sum_{k=0}^{\infty}\frac{1}{4^k}\\
&=\frac1{(2n+1)2^{2n+1}(1-1/4)}\\
&=\frac{4}{3(2n+1)2^{2n+1}}\\
&=\frac{1}{3(2n+1)2^{2n-1}}\\
\end{array}
$
Therefore,
you want
$\frac{1}{3(2n+1)2^{2n-1}}
\le 10^{-4}
$
or
$3(2n+1)2^{2n-1}
\ge 10^{4}
$.
Since
$2^{14} = 16384$,
start at $n=7$
and go down.
</s>