- #1
eyesontheball1
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Homework Statement
Hi, everyone. I'd appreciate it if someone could explain something for me regarding the convergence of series. Thanks in advance![/B]
Homework Equations
In my calculus book, I'm given the following:
(1) - For p > 1, the sum from n=1 to infinity of n^-p converges.
(2) - For the sum from n=1 to infinity of [(-1)^(n+1)]*(n^-p), if lim of n^-p approaches 0 as n approaches infinity and if (n+1)^-p <= n^-p, then this alternating series converges. It's clear that this series converges if p > 0.
So we have two series, series (1), which converges whenever p > 1, and series (2), which converges whenever p > 0. What I don't understand is why exactly I'm wrong in the following reasoning:
Suppose p > 1. p > 1 =>
sum{n=1, infinity}{n^-p} converges and 1-2^(1-p) converges =>
[1-2^(1-p)]*sum{n=1, infinity}{n^-p} converges, and
[1-2^(1-p)]*sum{n=1, infinity}{n^-p} = sum{n=1, infinity}{n^-p} - 2*sum{n=1, infinity}{(2n)^-p} =
sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p)} =>
[1-2^(1-p)]*sum{n=1, infinity}{n^-p} = sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p)} converges, but
sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p)} converges whenever p > 0 =>
[1-2^(1-p)]*sum{n=1, infinity}{n^-p} = sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p)} converges whenever p > 0, and
1-2^(1-p) /= 0 whenever p /= 1 =>
sum{n=1, infinity}{n^-p} = sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p)}/[1-2^(1-p)] converges whenever p > 0, p /= 1,
but we already know that sum{n=1, infinity}{n^-p} only converges for p s.t. p > 1, thus, we've arrived at a contradiction.