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.

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- December 11th 2008, 11:20 PMthahachainaradius 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. - December 12th 2008, 10:23 AMThePerfectHacker
Given a sequence the series converges if and only if where .

Therefore, converges if and only if where is chosen so large that .

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 then is on the open interval of convergence of and so converges too. But we have that and so converges too.