How do you find the maclaurin series expansion of f(x)=ln1+x5?

1 Answer
Jul 31, 2018

ln1+x5=n=0(1)nx5n+5n+1

with radius of convergence R=1.

Explanation:

Start from the geometric series:

11q=n=0qn

having radius on convergence R=1.
Let now q=x5:

11+x5=n=0(x5)n=n=0(1)nx5n

Still with R=1 because x51|x|1
Multiply by x4:

x41+x5=n=0(1)nx5n+4

In the interior of the interval of convergence we can integrate term by term obtaining a series with at least the same radius of convergence:

x0t41+t5dx=n=0(1)nx0t5n+4dt

15x0d(1+t5)1+t5dx=n=0(1)n[t5n+55n+5]x0

15ln1+x5=n=0(1)nx5n+55n+5

ln1+x5=n=0(1)nx5n+5n+1