# Proof involving Abel's Theorem

• December 8th 2009, 07:46 AM
Proof involving Abel's Theorem
Let lim n-> ∞ a_n = L. Then, let f(x) = ∑ from 0 to ∞ of (a_n)(x^n). Show that the lim x-> 1 (1-x)f(x) = L.

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.
• December 9th 2009, 04:42 AM
chisigma
Is...

$f(x)= \sum_{n=0}^{\infty} a_{n}\cdot x^{n}$ (1)

... so that...

$(1-x)\cdot f(x)= a_{0} + (a_{1}-a_{0})\cdot x + (a_{2} - a_{1}) \cdot x^{2} + \dots + (a_{n} - a_{n-1})\cdot x^{n} + \dots$ (2)

It is evident that (2) is a 'telescopic series' so that is...

$\lim_{x \rightarrow 1} (1-x)\cdot f(x) = \lim_{n \rightarrow \infty} a_{n} = L$ (3)

Kind regards

$\chi$ $\sigma$