How to find the sum of the infinite series?

Question

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!

Answer 1

Here is a detailed way to find the answer. Hopefully, that'll give you some insight you can use for similar questions.

1. $\frac{1}{2}$ 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 $f(1/2)$.
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?)
3. 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.)
4. But we know how to compute that derivative! For $x\in(-1,1)$, this is the sum of a geometric series: $$ g'(x) = \sum_{n=0}^{\infty} (x^2)^{n} = \frac{1}{1-x^2}\tag{4} $$
5. 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), $$ g(x) = g(0) + \int\frac{1}{1-x^2} = 0 + \frac12 \log\frac{1+x}{1-x}\tag{5} $$
6. We are done: $$ f(1/2) = {g(1/\sqrt2)}\cdot{\sqrt{2}} = \boxed{\frac1{\sqrt{2}} \log(3+2\sqrt2)} \approx 1.246 $$

Answer 2

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}

Answer 3

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).