Question

Absolute Convergence

"determine the range of values of x for which the corresponding series is absolutely convergent and evaluate its sum. You may use any theoretical results from the course without justification."

\[\sum_{n=0}^{\infty} \frac{ 2^{2n}x ^{n+1} }{ n! }\]

Answer 1

We know that sum x^n/n! converges to e^x. What if you subtracted your series above from this sum? I get that if x > 1/4 this series converges.

Answer 2

How do we know that it converges to e^x?

Answer 3

sum x^n/n! is the Taylor Series expansion of e^x around x=0. See http://en.wikipedia.org/wiki/Taylor_series

Answer 4

Oh I see!

Answer 5

When I subtract, I want to make sure that my series is actually less than e^x, because I know e^x converges, so x < 1/4 should have been my estimate. I wonder if you can use the root test or ratio test.

Answer 6

I tried using the ratio test earlier but I got that it diverges
(I'm new to this convergence stuff so I'm still trying to get my head around it)

Can you only use the convergence tests for series, or do they apply for sequences as well?

Answer 7

No, because, for example, the sequence 1/n converges by the ratio test but the series sum 1/n from 1 to inf converges.

Answer 8

I still think you can use the series for e^x. This is how...

Answer 9

Using the fact that \[e^{ax}=\sum_{n=0}^{\infty}\frac{(ax)^{n}}{n!}\]
We find that
\[\sum_{n=0}^{\infty}\frac{ 2^{2n}x ^{n+1} }{n!}\]
\[=\sum_{n=0}^{\infty}\frac{(2^{2})^{n}xx^{n}}{n!}\]
\[=\sum_{n=0}^{\infty}\frac{4^{n}xx^{n}}{n!}\]
\[=x\sum_{n=0}^{\infty}\frac{(4x)^{n}}{n!}\]
\[=xe^{4x}\]
Hence, the series converges for all x and converges to this function.

HTH

Answer 10

Also I get by the ratio test that \[\lim_{n \rightarrow \infty} \frac{ a _{n+1} }{ a_{n}}\]
\[=\lim_{n \rightarrow \infty}\frac{ 4x }{ n+1 }\]
\[=0\]
This shows that the underlying series converges, but that won't help you for this problem.

Answer 11

(sequence, not series)

Answer 12

Hi, thanks so much for being patient with me on this :)

I have followed your ratio rest [I get that now!], and I understand your working to get to
\[\sum_{n=0}^{\infty} \frac{ 2^{2n}x ^{n+1} }{ n! } = xe ^{4x}\]
but i don't understand how we know that
\[xe^{4x}\] converges?

(Sorry if this is really obvious!)

Answer 13

The series converges, because for every value of x, the series converges to xe^(4x) and nothing else (i.e. it's unique and doesn't wobble around, so to speak). It doesn't mean that there aren't some values of x which makes this function infinite, just that the sum converges to a definite value for every x < infinity. Another way to think about this is if for every x, you were to subtract the series from xe^(4x), then this error would tend to zero.

Answer 14

The error would tend to zero as you added more terms to the series.

Answer 15

That's the meaning of convergence here.

Answer 16

Oh I see! Thankyou so much for your help!

Answer 17

not a problem

Answer 18

BTW, this function converges absolutely because xe^(4x) <= |xe^(4x)| (just take the positive values in the series, which converge).

Answer 19

Another, BTW, since you asked. The convergence tests for sequences work only to test for divergence of the corresponding series. That is, if the sequence converges (but not to zero), then the corresponding series diverges. You can not say if the sequence converges to zero that the corresponding series converges as well (think of the series with 1/n as coefficients, which tend to zero but the corresponding series does not converge).