Convergence of a Taylor Series: Finding the Values of x

ThatOneGuy45
Messages
4
Reaction score
0

Homework Statement


For this problem I am to find the values of x in which the series converges. I know how to do that part of testing of convergence but constructing the summation part is what I am unsure about.

I am given the follwing:
1 + 2x + \frac{3^2x^2}{2!} +\frac{4^3x^3}{3!}+ ...

Homework Equations


I looked up online about the taylor series expansion for ex because I noticed it looked familiar and compared it with the series

The taylor series for ex is:
1 + x + \frac{x^2}{2!} +\frac{x^3}{3!}+ ...=Ʃ^{∞}_{n=0}\frac{x^n}{n!}

The Attempt at a Solution


What I did was pretty much just put \frac{(n+1)^nx^n}{n!} and checked the terms to see if it works. It seems to work but I am just a bit unsure. I haven't worked with this stuff in a while so I just want to be sure if I did that part right.
 
Last edited:
Physics news on Phys.org
That looks correct. You should be able to see if it converges using the ratio test, just be careful with the limit.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top