# Finding radius of convergence for series

Find the radius of convergence of the given power series:

$$\sum _{n=1}^{ \infty} \frac{(-1)^n x^{5n+2}}{5n+2}$$

Through the ratio test, I can get it down to $$\frac{(-1)^n+1 x^{5(n+1)+2}}{5(n+1)+2} /\frac{(-1)^n x^{5n+2}}{5n+2},$$ but then I am stuck.

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What have you tried? What do you know about how to find the radius of convergence of a power series? – Thomas Andrews Nov 8 '12 at 17:07
"Ctrl", unlike other sites, we typically require some effort on the part of the asker. Please edit your question to include your thoughts/attempts. You won't get better at math by having other people do the heavy lifting! – The Chaz 2.0 Nov 8 '12 at 17:14

For the radius of convergence, we have $$R=\lim_{n\to\infty}\left| \frac{(-1)^{n+1}x^{5n+7}}{5n+7}\frac{5n+2}{(-1)^nx^{5n+2}} \right| \\ =\lim_{n\to\infty}\left| \frac{(-1)x^5(5n+2)}{5n+7} \right| \\ =|x|^5\lim_{n\to\infty}\frac{5n+2}{5n+7} \\ =|x|^5\lim_{n\to\infty}\frac{5+\frac{2}{n}}{5+\frac{7}{n}}=|x|^5.$$ We will have convergence only if $|x|^5<1.$
Well, your numerator appears to have an addition sign that shouldn't be there. When I use the ratio test, I like to split the fraction into two, it's significantly easier to see what cancels out in this form. So: $lim_{n \rightarrow \infty} \Large|\frac{a_{n+1}}{1} \cdot \frac{1}{a_n}|$
$a_n$ is just going to be your general term. To find $a_{n+1}$, plug in n+1 into every instance of the variable n, in your general term. The key to simplifying what is inside the absolute values, to easily find the limit, is being proficient with exponents.