Have you tried substituting in x = -1 and x = 1? Obviously the series must be divergent when x = -1 because the logarithm is undefined at 0. As for when x = 1, we then have , which is the alternating harmonic series and is known to be convergent.
I can do a, b, c and d alright, but I'm stuck on e.(a) Write down the Maclaurin series for . What it it's radius of convergence?
(b) Write down the Maclaurin series for
(c) Use (b) to find the Maclaurin series for f(x) = ln (1+x)
(d) What is the radius of convergence found in (c)?
(e) What is the interval of convergence found in (c)?
(a)
The radius of convergence is 1 (It's a geometric series where x is the ratio).
(b)
(c)
(d) The radius of convergence is still 1, integrating does not alter the radius of convergence.
Now how do I go about checking the interval of convergence? Is it just (-1, 1)?
Have you tried substituting in x = -1 and x = 1? Obviously the series must be divergent when x = -1 because the logarithm is undefined at 0. As for when x = 1, we then have , which is the alternating harmonic series and is known to be convergent.
Oh right, haha, so the interval of convergence is (-1, 1]
How would I know that the interval of convergence is around 0? Is there any series (can you give an example?) where the interval of convergence is centered around something else?