# radius of convergence (power series)

Printable View

• Dec 11th 2008, 10:20 PM
thahachaina
radius of convergence (power series)
Hey guys,

if |a_k|<|b_k| for large k, show that if (b_k)(x^k) converges on an open interval I, then (a_k)(x^k) also converges on I.

Please help.
• Dec 12th 2008, 09:23 AM
ThePerfectHacker
Quote:

Originally Posted by thahachaina
Hey guys,

if |a_k|<|b_k| for large k, show that if (b_k)(x^k) converges on an open interval I, then (a_k)(x^k) also converges on I.

Please help.

Given a sequence $\{ x_n\}$ the series $\sum_{n=1}^{\infty}x_n$ converges if and only if $\sum_{n=N}^{\infty} x_n$ where $N>1$.
Therefore, $\sum_{k=1}^{\infty}a_kx^k$ converges if and only if $\sum_{k=K}^{\infty}a_kx^k$ where $K>1$ is chosen so large that $|a_k| < |b_k|$.

Remember that an interval of converges of a series (centered as 0) is an interval centered at 0. If we look at the open interval then the series also converges absolutely. Now if $x\in I$ then $x$ is on the open interval of convergence of $\sum_{k=K}^{\infty}b_kx^k$ and so $\sum_{k=K}^{\infty}|b_kx^k|$ converges too. But we have that $|a_kx^k| \leq |b_kx^k|$ and so $\sum_{k=K}^{\infty}a_kx^k$ converges too.