Is the Convergence of this Infinite Series Dependent on the Value of Alpha?

[JG89]

Homework Statement

For what values of alpha is the following series convergent: \sum_{v=1}^{\infty} \frac{(-1)^{n-1}}{n^{\alpha}} = 1 - \frac{1}{2^{\alpha}} + \frac{1}{3^{\alpha}} + ...

Homework Equations

The Attempt at a Solution

For negative alpha and alpha = 0 the series obviously diverges, so we need only look at positive alpha. For alpha = 1, the infinite series for ln(2) arises. For alpha > 1, we see that the series is absolutely convergent and so the original series is also convergent.

Now to prove that the series is divergent for 0 < alpha < 1, notice that -1/n \le \frac{(-1)^{n-1}}{n^{\alpha}} and so \sum_{v=0}^{n} -1/n \le \sum_{v=0}^{n} \frac{(-1)^{n-1}}{n^{\alpha}}.

\frac{(-1)^{n-1}}{n^{\alpha}} = -1 - 1/(2^{\alpha}) - 1/(3^{\alpha}) - ... - 1/(n^{\alpha}) = -(1 + 1/2^{\alpha} + ... + 1/(n)^{\alpha} ). The series in the brackets diverges for increasing n since alpha is less than 1, and so the entire series converges.

Thus the original series converges only for alpha >= 1.

Is this correct?

 

[JG89]

Ah, I messed up my latex and I can't edit it. The inequality should actually be the sum from 0 to n of -1/n^alpha <= sum from 0 to n of {(-1)^(n-1)}/n^alpha

And the series the series on the left hand side of the inequality is -(1 + 1/2^alpha + 1/3^alpha + ...), which diverges since 0 < alpha < 1

 

[Dick]

Review the alternating series test.

 

[JG89]

The test says that if the absolute value of the terms of an alternating series decrease monotonically to zero then the series converges.

So let a_n be the nth term of the series. Then |a_n| = 1/n^alpha > |a_{n+1}| = 1/(n+1)^alpha

The terms do not decrease monotonically to 0, but this doesn't mean that the series doesn't converge...so I'm still stuck...

 

[Office_Shredder]

So let a_n be the nth term of the series. Then |a_n| = 1/n^alpha > |a_{n+1}| = 1/(n+1)^alpha

The terms do not decrease monotonically to 0, but this doesn't mean that the series doesn't converge...so I'm still stuck...

Review the definition of decreasing monotonically to zero

 

[JG89]

Ha! I can't believe I made that mistake...the absolute value of the terms indeed do go to 0 and so the series converges

 

[Dick]

You say |a_n|>|a_{n+1}|. That certainly looks like you are saying the series decreases. But that's not true for every alpha. Certainly not if, for example, alpha=(-1). For which alpha is that true?

 

[JG89]

It is true for positive alpha

 

[Dick]

Right.