Uniform convergence of series question?

Question

If the sum (n=1 -> infinity) a_n < infinity, then show that sum (n=1->infinity) a_n*x^n converges uniformly on 0<=x<=1.

Since 0<=x<=1, obviously the second series < infinity as well. I also know, by a theorem, that if you can show that a series converges at 0 and 1, means that the entire interval is also uniformly convergent.

R = lim (n-> infinity) |a_n+1/a_n| = 1, if i did that right. And |x|<=1 as well, then it converges at 0 and at 1, so there it is uniformly convergent on [0,1].

Am I close?

kb · Accepted Answer

If R = 1, that tells us nothing about the convergence of a series.

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Let ε > 0 be given.

Since Σ(n=1 to ∞) a_n converges, the Cauchy Criterion for convergent series guarantees

that there exists a positive integer N such that for all n > m ≥ N we have

|a_(m+1) + a_(m+2) + ... + a_n| < ε/4.

Let s_n = a_(m+1) + a_(m+2) + ... + a_n.

Note that |s_n| < ε/4.

So for all x in [0, 1],

|a_(m+1) x^(m+1) + a_(m+2) x^(m+2) + ... + a_n x^n|

= |Σ(j=m+1 to n) a_j x^j|

= |s_n x^(n+1) - s_m x^(m+1) + Σ(j=m+1 to n) s_j (x^j - x^(j+1))|, since a_n = s_n - s_(n-1)

≤ |s_n| x^(n+1) + |s_m| x^(m+1) + Σ(j=m+1 to n) |s_j| |x^j - x^(j+1)|

< (ε/4) [x^(n+1) + x^(m+1) + Σ(j=m+1 to n) (x^j - x^(j+1))]

= (ε/4) [x^(n+1) + x^(m+1) + (x^(m+1) - x^(n+1)], via telescoping sums.

= (ε/4) [2x^(m+1)]

≤ (ε/4) * 2 * 1^(m+1)

= ε/2

< ε.

Hence, Σ(n=1 to ∞) a_n x^n converges uniformly on [0, 1] by Cauchy Criterion.

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P.S.: I essentially reproved Abel's Lemma (via partial summation) in the chain of inequalities above for completeness.

I hope this helps!

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Uniform convergence of series question?

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