Summation by Parts: Lim x->1 (1-x)f(x)=L

[tracedinair]

Homework Statement

Let lim n-> ∞ a_n = L. Then, let f(x) = ∑ from n=0 to ∞ of (a_n)(x^n). Show that the lim x-> 1 (1-x)f(x) = L.

Homework Equations

The Attempt at a Solution

This one is pretty far over my head. I know at some point you're supposed to use Abel/SBP, but here is what I have so far.

Let |a_n| go to |L|.

Then, using the ratio test, let |a_n|^(1/n) go to |L|^(1/∞) = |L|^(0) = 1.

Then, from the Ratio test, we can see that the series will converge for |x| < 1.

Take ∑ from 0 to ∞ of (a_n)(x^n). Then, multiply through. So, we obtain, (1-x)∑ from 0 to ∞ of (a_n)(x^n) = ∑ from 0 to ∞ of (x^n - x^(n+t)). Taking b_n to equal (x^n - x^(n+t)) we can get ∑ (a_n)(b_n)..

This is where I get stuck. I'm not really where to take it from here.

 

[Billy Bob]

Here's my way. Start over. (You won't use Abel or SBP.)

You want to show |(1-x)f(x)-L| is small.

Use a "trick" that L=(1-x)L/(1-x) and write 1/(1-x) as the geometric series.

This enables you to express |(1-x)f(x)-L| as a single power series.

Since a_n approaches L, we have |a_n - L| < epsilon, or epsilon/2, or whatever you need, for n>N.

Break your power series for |(1-x)f(x)-L| into two summations, one up to N, the other for N+1 and beyond.

For the summation up to N, this is continuous at x=1.

For the summation starting at N+1, get an upper bound. You will again use the geometric series.

I hope I didn't give away too much or too little. This plan comes from looking at the proof of Abel's Theorem. In that proof, essentially one uses SBP to reduce the given problem to your problem.