ind the radius of convergence, R,and the interval, I, of the series.
n=3
[x^(n+1)]/(5n!)
a good place to start is this site, with several useful examples. See what you make of it.
http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx
To find the radius of convergence (R) and the interval (I) of the series, we need to use the ratio test. The ratio test checks for convergence by taking the limit of the absolute value of the ratio of consecutive terms in the series.
Let's apply the ratio test to the given series:
a_n = [x^(n+1)]/(5n!)
To find the limit as n approaches infinity of the ratio of the (n+1)th term to the nth term, we need to calculate:
lim(n->∞) |((x^(n+2))/(5(n+1)!))/(x^(n+1)/(5n)!)|
The ratio of the (n+1)th to nth terms simplifies to:
lim(n->∞) |x^(n+2)/(x^(n+1))| * |(5n)/(5(n+1))|
Simplifying further:
lim(n->∞) |x| * |[(5n)/(5n+5)]|
As n approaches infinity, the term (5n)/(5n+5) simplifies to 1.
So, the ratio test simplifies to:
lim(n->∞) |x|
We want this limit to be less than 1 for convergence.
|x| < 1
This inequality indicates that the series converges for values of x that satisfy |x| < 1. By definition, the distance from 0 to the boundary of this interval is the radius of convergence (R).
Therefore, the radius of convergence (R) is 1.
The interval (I) of convergence is the open interval of x-values between -R and R. In this case, it is the open interval (-1, 1).