Question

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

Answer 1

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->oo) (-1(n + 1)^(n + 1)x)/((n + 1)n^n) < 1`

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

Let's consider the limit as `n -> oo` of `((n + 1)/n)^n`. This is a standard limit known as being `e`

Thus

`|x|(e) < 1`

`|x| < 1/e`

So our interval of convergence will be

`(-1/e,1/e)`

However, we must test endpoints. When `x= -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

`1/e + 2/e^2 + 9/(2e^3) + 32/(3e^4) + ...`

You can use the root test to see that this converges . As for `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

`-1/e + 2/e^2 - 9/(2e^3) + 32/(3e^4) + ...`

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

Our interval of convergence is therefore `[-1/e, 1/e]`, and our radius is `1/e`.

Hopefully this helps!