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
 A: $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.
A: Hint: 
$$
\sum_{k=0}^\infty \frac{1}{2^{2k+1}} = \frac{2}{3}
$$
and by scaling by an appropriate power of $1/2$ you can also find 
$$
\sum_{k=j}^\infty \frac{1}{2^{2k+1}}
$$
for any $j$
A: Hint:
$\sum_{k=0}^\infty\frac{x^{2k+1}}{2k+1} $  is the Taylor series of the function 
$$\operatorname{argtanh}(x)=\frac 12\ln\frac{1+x}{1-x}\qquad\text{(radius of convergence: $1$).}$$
Indeed, if you've never seen this function, its derivative  is equal to $\;\dfrac 1{1-x^2}$.
A: Start from the following:
$\sum\limits_{k=0}^{\infty}\frac{1}{2^{2k+1}(2k+1)}+\sum\limits_{k=1}^{\infty}\frac{1}{2^{2k}2k}=\sum\limits_{n=1}^{\infty}\frac{1}{2^{n}n}\tag1$
$n=2k$, if n even
$n=2k+1$, if n odd
$S=\sum\limits_{k=1}^{\infty}\frac{1}{2^{2k+1}(2k+1)}=\sum\limits_{n=1}^{\infty}\frac{1}{2^{n}n}-\frac{1}{2}\sum\limits_{k=1}^{\infty}\frac{1}{4^{k}k}=Li_1(\frac{1}{2})-\frac{1}{2}Li_1(\frac{1}{4})\tag2$
where $Li_1(z)$ is the polylogarithm function.
Known that $Li_1(z)=-\ln(1-z)$, based on the result of (2) we get the value of the series:
$S=\ln2+\frac{1}{2}\ln(\frac{3}{4})=\frac{1}{2}\ln3\tag3$
So 
$\sum\limits_{k=0}^{\infty}\frac{1}{2^{2k+1}(2k+1)}=\frac{1}{2}\ln3\tag4$
A: Just for your curioisity.
Starting from marty cohen' answer, and making the problem more general, you need to solve for $n$
$$\frac{1}{3(2n+1)2^{2n-1}}=\epsilon$$ the solution is given by
$$n=\frac{1}{2 \log (2)}W\left(\frac{4 \log (2)}{3 \epsilon }\right)-\frac{1}{2}$$ where $W(t)$ is Lambert function. For sure, you will need to use $\lceil n\rceil$.
Since the argument is large, you can use the approximation given in the linked page
$$W(t)=L_1-L_2+\frac{L_2}{L_1}+\cdots \quad \text{where} \quad L_1=\log(t)\quad \text{and} \quad L_2=\log(L_1)$$
Applied to you case $(\epsilon=10^{-4})$, this will give as a real $n=4.67$, then $5$ is your answer.
