A Solving a power series

[ILY]

The question was: calculate the following sum within its open interval of convergence after determining the radius of convergence:

$\sum_{n=0}^{+\infty} \frac{x^n}{2n+1}$

$\textbf{Finding the radius of convergence:}$

I believe I followed the correct steps, but I got stuck solving it. Here's what I've done so far:

We use the ratio test to determine the radius of convergence. The formula for the radius of convergence, R, is:

$\frac{1}{R} = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$

In this case, the general term $a_n = \frac{x^n}{2n+1}$. Applying the ratio test:

$\frac{1}{R} = \lim_{n \to \infty} \left| \frac{\frac{x^{n+1}}{2(n+1)+1}}{\frac{x^n}{2n+1}} \right|$

Simplifying:

$\frac{1}{R} = \lim_{n \to \infty} \left| \frac{x^{n+1}(2n+1)}{(2n+3)x^n} \right|$
$\frac{1}{R} = \lim_{n \to \infty} \left| \frac{x(2n+1)}{2n+3} \right|$

Now, as $n \to \infty$, $\frac{2n+1}{2n+3} \to 1$. Therefore:

$\frac{1}{R} = |x| \cdot 1 = |x|$

Thus, the series converges when |x| < 1, and the radius of convergence is R = 1.

I think this is correct, but I got stuck while solving it. If anyone could help me, it would be greatly appreciated!

 

[Mark44]

Fixed all of your LaTeX, which was technically correct but used incorrect delimiters. Here at PF the delimiters are ##, for inline LaTeX, and $$, for standalone LaTeX. See the LaTeX Guide at the link in the lower left corner of this pane.

I think this is correct, but I got stuck while solving it.

I don't see anything wrong in the work you show.

By "solving it," I presume you mean finding the sum of the series. What are some of the techniques shown in your textbook for finding the sum of a convergent power series? Your textbook might have some examples.

 

[pasmith]

Consider <br /> \operatorname{arctanh}(x) = \sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}, \quad |x| &lt; 1. Now you just have to manipulate the power so that it is n rather than 2n+1. Can you see how to do that? I would start with 0 \leq x &lt; 1, and then try ot extend the result to -1 &lt; x &lt; 0.

 

[pasmith]

Alternatively, if <br /> f(x) = \sum_{n=0}^\infty \frac{x^n}{2n+1} then <br /> f&#039;(x) = \sum_{n=0}^\infty \frac{nx^{n-1}}{2n+1} so that <br /> 2xf&#039;(x) + f(x) = \sum_{n=0}^\infty x^n = \frac{1}{1 - x} is a differential equation which can be solved for f subject to f(0) = 1.