Radius of convergence and interval of convergence

Quote:

(a) Write down the Maclaurin series for $\displaystyle \dfrac{1}{1-x}$. What it it's radius of convergence?

(b) Write down the Maclaurin series for $\displaystyle \dfrac{1}{1+x}$

(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)?

I can do a, b, c and d alright, but I'm stuck on e.

(a) $\displaystyle \dfrac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + x^5 ...$

The radius of convergence is 1 (It's a geometric series where x is the ratio).

(b) $\displaystyle \dfrac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 - x^5 ...$

(c) $\displaystyle \displaystyle \int \dfrac{1}{1+x} = \ln (1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \dfrac{x^5}{5} - \dfrac{x^6}{6} ...$

(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)?

Re: Radius of convergence and interval of convergence

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 $\displaystyle \displaystyle \begin{align*} 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots - \dots \end{align*}$, which is the alternating harmonic series and is known to be convergent.

Re: Radius of convergence and interval of convergence

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?

Re: Radius of convergence and interval of convergence

A MacLaurin Series is centred at 0 and so therefore is its radius of convergence, a general Taylor series centred at x = c will have a radius of convergence centred at x = c.