How to find the sum of the infinite series? So I just started learning series, and I am having trouble understanding them. I have a series below: 
$$\sum_{n=0}^{\infty} \frac{1}{(2n+1)2^n}$$
Can someone explain to me how I should go about this, from start to finish? I couldn't find any online websites that explained how to find sums, so if you have that, it would be greatly appreciated!
 A: Here is a detailed way to find the answer. Hopefully, that'll give you some insight you can use for similar questions.


*

*$\frac{1}{2}$ is just a number; your series is just a number. To rely on the whole power and flexibility of real analysis, functions are more useful. The first trick is to define the function
$$
f(x) = \sum_{n=0}^{\infty} \frac{x^n}{2n+1}\tag{1}
$$
which is defined for all $x\in(-1,1)$. (Can you argue why?)
Then what you want to compute is $f(1/2)$.

*Why does it help? So far, it's not clear, but what is inside the sum (i.e., the summands) kind of looks like a derivative.  In particular, if I look at
$$
g(x) = x f(x^2) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}\tag{2}
$$
then it should be clear that the $n$-th summand looks like the antiderivative (with regard to $x$) of $x^{2n}$. Also, note that the original problem is now to compute $g(1/\sqrt{2})\cdot\sqrt{2}$, so we're not losing track of the end goal. (Note: $g$ is also defined on $(-1,1)$: do you see why?)

*Here is the fun part. One can actually argue, using theorems about power series within their radius of convergence, that the above eyeballing is justified. Namely, we do have
$$
g'(x) = \frac{d}{dx} \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}
= \sum_{n=0}^{\infty} \frac{d}{dx} \frac{x^{2n+1}}{2n+1}
= \sum_{n=0}^{\infty} x^{2n}\tag{3}
$$
(The tricky part here is arguing that swapping $\sum$ and $\frac{d}{dx}$ is alright.)

*But we know how to compute that derivative! For $x\in(-1,1)$, this is the sum of a geometric series:
$$
g'(x) 
= \sum_{n=0}^{\infty} (x^2)^{n}
= \frac{1}{1-x^2}\tag{4}
$$

*We're almost done. All that remains is to... integrate that explicit derivative we found to get back the expression for $g$: recalling from the expression in (2) that $g(0)=0$ (to help us figure out the constant of integration),
$$
g(x) = g(0) + \int\frac{1}{1-x^2} = 0 + \frac12 \log\frac{1+x}{1-x}\tag{5}
$$

*We are done:
$$
f(1/2) = {g(1/\sqrt2)}\cdot{\sqrt{2}} = \boxed{\frac1{\sqrt{2}} \log(3+2\sqrt2)} \approx 1.246
$$
A: Using inverse hyperbolic tangent's Maclaurin series $\text{arctanh} x=\sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1},$
\begin{equation}
\begin{split}
\sum_{n=0}^{\infty} \frac{1}{(2n+1)2^n}&=\sqrt2\sum_{n=0}^{\infty} \frac{1}{2n+1}\left(\frac1{\sqrt 2}\right)^{2n+1}\\
&=\sqrt2\text{arctanh}\left(\frac1{\sqrt2} \right)\\
&=\sqrt2\cdot\frac12\left(\ln\left(1+\frac1{\sqrt2}\right)-\ln\left(1-\frac1{\sqrt2}\right)\right)\\
&=\sqrt2\ln(1+\sqrt2)\\
\end{split}
\end{equation}
A: Try factoring the argument as 1/(2n+1)*1/2^n. Then do backwards long division on the 1st fraction I wrote. This should allow you to condense the entire thing into a geometric series. From there you can use S=a/(1-r).
