Convergence..

[Daniel_Feldman]

Can someone tell me why, between the alternating series test and the p-series test, I'm getting stuck?


I need to determine the values of p for which the series from 1 to infinity, of (-1)^(n-1)/n^p will converge.

I know that for the p-series 1/n^p, it converges for p>1. This one, however, looks like an alternating series, and somewhere I'm confusing myself...I think.

 

[soroban]

Hello, Daniel!

Can someone tell me why, between the alternating series test and the p-series test, I'm getting stuck?

It can be confusing . . . it took me some time to wake up.

Determine the values of p for which the series \(\displaystyle \L\sum^{\infty}_{n=1}\,\frac{(-1)^{n-1}}{n^p}\) will converge.

Since it is an alternating series, it's a bit easier.

An alternating series converges if the terms go to zero.

When does $\displaystyle \,\frac{1}{n^p}\,\to\,0$ ? . . . It goes to 0 for any $\displaystyle p\,>\,0$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I had to baby-talk my way through it . . .

We know it coverges for $\displaystyle p\,>\,1$ . . . but what about: $\displaystyle \,0\,<\,p\,\leq\,1$

When $\displaystyle p\,=\,1$, we have the alternating harmonic series . . . which converges.

Consider $\displaystyle \,p\,=\,\frac{1}{9}$, then the general term is: \(\displaystyle \L\,\frac{1}{\sqrt[9]{n}}\) . . . and it goes to 0.

Even if $\displaystyle p\,=\,\frac{1}{1000}$, the general term goes to 0.


Note: we do not want $\displaystyle p\,=\,0$.
The series becomes: $\displaystyle \sum (-1)^{n-1} \:=\:1\,-\,1\,+\,1\,-\,\cdots\:$ which diverges.

 

[pka]

You may want to consider absolute convergance.

 

[Daniel_Feldman]

Hello, Daniel!

Can someone tell me why, between the alternating series test and the p-series test, I'm getting stuck?

It can be confusing . . . it took me some time to wake up.

[quote:2fp6skwc]Determine the values of p for which the series \(\displaystyle \L\sum^{\infty}_{n=1}\,\frac{(-1)^{n-1}}{n^p}\) will converge.

Since it is an alternating series, it's a bit easier.

An alternating series converges if the terms go to zero.

When does $\displaystyle \,\frac{1}{n^p}\,\to\,0$ ? . . . It goes to 0 for any $\displaystyle p\,>\,0$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I had to baby-talk my way through it . . .

We know it coverges for $\displaystyle p\,>\,1$ . . . but what about: $\displaystyle \,0\,<\,p\,\leq\,1$

When $\displaystyle p\,=\,1$, we have the alternating harmonic series . . . which converges.

Consider $\displaystyle \,p\,=\,\frac{1}{9}$, then the general term is: \(\displaystyle \L\,\frac{1}{\sqrt[9]{n}}\) . . . and it goes to 0.

Even if $\displaystyle p\,=\,\frac{1}{1000}$, the general term goes to 0.


Note: we do not want $\displaystyle p\,=\,0$.
The series becomes: $\displaystyle \sum (-1)^{n-1} \:=\:1\,-\,1\,+\,1\,-\,\cdots\:$ which diverges.[/quote:2fp6skwc]


Thanks soroban...quick question though.

For this alternating series (by definition), it converges if the limit as n approaches infinity of 1/n^p=0 and if 1/n^p is always decreasing. How do you go from that to "it converges if its terms go to 0"?


Maybe I'm just not thinking straight right now....