1. ## Radius of convergence..Taylor series..

Hey guys..I was revising and stumbled onto this question which the lecturer talked about..but I cant seem to get to how he got the Radius of convergence...heres the question..

Find the Taylor series for f(x)=e^x about x=1..from this use the ratio test to find the radius of convergency...

could someone please show me how this is done..Thanks..

2. The kth term is x^k/(k!)

lim(ck+1)/ck =

lim|x^(k+1)*k!/[(x^k*(k+1)!]|= |x| lim1/(k+1) < 1

|x|< lim(k+1) = infinity

So the radius of convergence is inf and the IC (-inf,inf)

3. Originally Posted by Calculus26
The kth term is x^k/(k!)

lim(ck+1)/ck =

lim|x^(k+1)*k!/[(x^k*(k+1)!]|= |x| lim1/(k+1) < 1

|x|< lim(k+1) = infinity

So the radius of convergence is inf and the IC (-inf,inf)
But the expansion is about $\displaystyle x= 1$ so the kth term would be

$\displaystyle \frac{e(x-1)^k}{k!}$

The interval of convergence is still all real x.

4. good catch---didn't even notice the expansion was about x = 1