# Find the interval and radius of convergence of the following power series (problem 1a)?

## May 12, 2018

Use the ratio test to find the radius of convergence. You can then determine the interval from there.

lim_(n->oo) ((-1)^(n + 1)(n + 1)^(n + 1)x^(n + 1))/(((n + 1)!)/(((-1)^n n^nx^n)/(n!))) < 1

${\lim}_{n \to \infty} \frac{- 1 {\left(n + 1\right)}^{n + 1} x}{\left(n + 1\right) {n}^{n}} < 1$

${\lim}_{n \to \infty} \frac{- 1 {\left(n + 1\right)}^{n} x}{{n}^{n}} < 1$

Let's consider the limit as $n \to \infty$ of ${\left(\frac{n + 1}{n}\right)}^{n}$. This is a standard limit known as being $e$

Thus

$| x | \left(e\right) < 1$

$| x | < \frac{1}{e}$

So our interval of convergence will be

$\left(- \frac{1}{e} , \frac{1}{e}\right)$

However, we must test endpoints. When $x = - \frac{1}{e}$, we get:

(-1)^n (n^n (-1/e)^n)/(n!)

We can start by seeing that the negative bases will cancel and always give positives, so we can rewrite

(n^n (1/e)^n)/(n!)

Now we can write the first few terms down

$\frac{1}{e} + \frac{2}{e} ^ 2 + \frac{9}{2 {e}^{3}} + \frac{32}{3 {e}^{4}} + \ldots$

You can use the root test to see that this converges . As for $\frac{1}{e}$, we repeat the same process, however this time we will have an alternating series.

(-1)^n (n^n(1/e)^n)/(n!)#

Write the first few terms down

$- \frac{1}{e} + \frac{2}{e} ^ 2 - \frac{9}{2 {e}^{3}} + \frac{32}{3 {e}^{4}} + \ldots$

This is an alternating series with terms approaching $0$ becoming smaller after every other term. Thus it converges.

Our interval of convergence is therefore $\left[- \frac{1}{e} , \frac{1}{e}\right]$, and our radius is $\frac{1}{e}$.

Hopefully this helps!