Need help!!

[quiza]

Find the radius of convergence and interval of convergence

of the power series

∞

∑ 2^n . x^n
n=1 n!

Ignore the problem of convergence

at the end points of the interval.

[Dingus]

I think this should be in the GRE-Math forum.

[Holden_Caulfield]

Originally Posted by Dingus

I think this should be in the GRE-Math forum.

Dingus, I should say that I disagree with you
Here is the right place for this, because dealing with this problem requires knowledge beyond the high-school level (hence, in my humble opinion).

[Naseer Asif]

quiza,

Here, a(n)=2^n.x^n/n! => a(n+1)=2^(n+1).x^(n+1)/(n+1)!
So, |a(n+1)/a(n)|=|2x/(n+1)|
For convergence cosider limit x->infinity|a(n+1)/a(n)|=limit x->infinity|2x/(n+1)|=0
=>The series converges for all the +ve values of x
and radius of convergence is infinity

[Dingus]

Originally Posted by Holden_Caulfield

Dingus, I should say that I disagree with you
Here is the right place for this, because dealing with this problem requires knowledge beyond the high-school level (hence, in my humble opinion).

I totally agree. This thread was in the Lounge earlier. Perhaps I should have clarified that by GRE-Math, I meant GRE-(Subject-test)-Mathematics.

[Dragonfinity]

Originally Posted by Naseer Asif

Here, a(n)=2^n.x^n/n! => a(n+1)=2^(n+1).x^(n+1)/(n+1)!
So, |a(n+1)/a(n)|=|2x/(n+1)|
For convergence cosider limit x->infinity|a(n+1)/a(n)|=limit x->infinity|2x/(n+1)|=0
=>The series converges for all the +ve values of x
and radius of convergence is infinity

Correct, except consider the limit as n -> infinity.

[Dragonfinity]

also, the series converges for all negative values of x. without stating this, you would have a radius of convergence of (0, infinity), or [0, infinity).

[Naseer Asif]

Dragonfinity,

limit x->infinity was typing mistake,I am agree with you and infact it is limit n->infinity. But please explain how the series converges for -ve values too?

Thanks for rectification and comments

[Dragonfinity]

For ANY value of x, 2x is fixed (and thus bounded). Since 1/(n+1) -> 0 as n-> infinity and 2x is bounded, 2x/(n+1) -> 0 as n -> infinity.

C

[Naseer Asif]

Thanks, Dragonfinity