Question

Find the radius of convergence and the interval of convergence of the following infinite series:
\[\sum_{n=1}^{\infty}{\frac{n!*x^n}{1*3*5*...*(2n-1)}}\]

Answer 1

:-( I need help with this one...

Answer 2

are you sure it converges?

Answer 3

it must (on condition)... it's a power series.

Answer 4

I got it to be convergent for \(-2\le x \le 2\). You need to test it at the points \(x=-2\) and \(x=2\).

Answer 5

again...use the ratio test

Answer 6

\[\left| {a_{n+1} \over a_n} \right|=\left| {n!(n+1)x^{n+1} \over 1 \times 3 \times 5 \times .. \times (2n+1)}. {1 \times 3 \times 5 \times.. (2n-1) \over n!x^n} \right|=\left| {(n+1)x \over 2n+1} \right|.\]
\(\left| {n+1 \over 2n+1} \right|\rightarrow{1 \over 2} \text{ as n }\rightarrow \infty\). So it converges for \(-1 < \frac{1}{2}x< 1 \implies -2<x<2\).

Answer 7

how do I test the bounds of the interval?

Answer 8

You need now to test the points where \(|\frac{1}{2}x|=1\), that's x=2 and x=-2 using any other test.

Answer 9

what test would I use? I can't think of any :-(

Answer 10

Neither can I :D! I'm thinking!

Answer 11

obvious...use divergence test

Answer 12

Divergence test would work with \(x=2\), but not with \(x=-2\), right?

Answer 13

it works with both

Answer 14

if the limit is not zero or does not converge the series diverges

Answer 15

what's the divergence test? you mean simply testing the limit of the sequence? it looks inconclusive :-(

Answer 16

The limit is not \(0\) yeah! I can see that.

Answer 17

The divergence test states that if the limit of \(a_n\) is not zero, then the series must be divergent.