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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.