Radius and Interval of Convergence for $\sum_{n=1}^{\infty}\frac{5^n}{n^2}x^n$ $$\sum_{n=1}^\infty \frac{5^n}{n^2}x^n$$
After doing the ratio test I end up with:
$$5|x| < 1$$
I'm confused, though, as to what is considered my interval of convergence and what is my radius.  I recognize that $1$ is my limit, so does this mean my radius of convergence is $\frac{1}{5}$ and my interval of convergence is:
$$\left(-\frac{1}{5},\frac{1}{5}\right)$$
 A: From the ratio test you've shown that the interval of convergence may be $(-\frac{1}{5},\frac{1}{5})$.  To finish up plug in $x=\frac{1}{5}$ and $x=\frac{-1}{5}$.  The first is just, $$\sum_{n=1}^\infty \frac{5^n}{n^2}\frac{1}{5^n}=\sum_{n=1}^\infty \frac{1}{n^2}$$ which converges by p-series test.  So at the very least you can say the interval of convergence is $(-\frac{1}{5},\frac{1}{5}]$.  Can you finish it up from here?
A: Why not directly Cauchy' Hadamard formula with the $\;n\,-$ the roots test?
$$\sqrt[n]{\frac{5^n}{n^2}}=\frac5{\sqrt[n]{n^2}}\xrightarrow[n\to\infty]{}5\implies R=\frac15$$
is the convergence radius, so the series converges for $\;-\frac15<x<\frac15\;$ . You can now check that at both extreme points the series is also convergent, so the interval of convergence is $\;\left[-\frac15,\,\frac15\right]\;$
A: I wonder if you'd find it less confusing if it were written like this:
$$
\sum_{n=1}^\infty \frac{5^n}{n^2} (x- 23)^n
$$
(where the number $23$ was chosen somewhat arbitrarily).
Instead of $5|x|<1$ you'd have $5|x-23|<1$, which implies  $|x-23|<\dfrac 1 5$, and that is equivalent to $-\dfrac 1 5 < x-23 < \dfrac 1 5$, and finally to $$23-\dfrac 1 5 < x < 23+\dfrac 1 5.$$  Then what would you take to be the interval of convergence?
