1. ## Power Series #2

This is the last of the set of real analysis questions. Any help would be greatly appreciated:

A series $\displaystyle \sum_{n=0}^{\infty}{{a_n}}$ is said to be Abel-summable to L if the power series
f(x) = $\displaystyle \sum_{n=0}^{\infty}$ a_n*x^n converges for all x Є [0,1) and L = lim f(x) as x approaches 1 from the negative side.

a. Show that any series that converges to a limit L is also Abel-summable to L.
b. Show that $\displaystyle \sum_{n=0}^{\infty}$ (-1)^n is Abel-summable and find the sum.

2. Originally Posted by ajj86

A series $\displaystyle \sum_{n=0}^{\infty}{{a_n}}$ is said to be Abel-summable to L if the power series
f(x) = $\displaystyle \sum_{n=0}^{\infty}$ a_n*x^n converges for all x Є [0,1) and L = lim f(x) as x approaches 1 from the negative side.

a. Show that any series that converges to a limit L is also Abel-summable to L.
since $\displaystyle \sum a_n$ is convergent, the interval of convergence of the power series $\displaystyle \sum a_nx^n$ contains 1. thus the radius of convergence is at least 1, i.e. it converges at least over the interval $\displaystyle (-1,1].$

b. Show that $\displaystyle \sum_{n=0}^{\infty}$ (-1)^n is Abel-summable and find the sum.
suppose $\displaystyle |x| < 1.$ then $\displaystyle \sum_{n \geq 0} (-1)^n x^n = \sum_{n \geq 0} (-x)^n = \frac{1}{1+x}.$ therefore $\displaystyle L=\lim_{x\to1^{-}} \sum_{n \geq 0}(-1)^n x^n = \frac{1}{2}.$
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