# Convergence/Divergence of an infinite series

1. Oct 19, 2009

### utleysthrow

1. The problem statement, all variables and given/known data

$$\sum^{\infty}_{n=1} \frac{(-1)^{n}}{\sqrt{n}+(-1)^{n}}$$

Prove whether this series converges or diverges using the following analysis:
Let $$b_{k}$$ be the sum of terms numbered n=2k-1 and n=2k from the given series.
By simplifying $$b_{k}$$, determine if $$\sum^{\infty}_{k=1} b_{k}$$ converges or diverges.

2. Relevant equations

3. The attempt at a solution

From what I have above, I think:

$$b_{k} = \frac{(-1)^{2k-1}}{\sqrt{2k-1}+(-1)^{2k-1}} + \frac{(-1)^{2k}}{\sqrt{2k}+(-1)^{2k}}$$

which means,

$$\sum^{\infty}_{k=1} b_{k}= \sum^{\infty}_{k=1} \frac{(-1)^{2k-1}}{\sqrt{2k-1}+(-1)^{2k-1}} + \sum^{\infty}_{k=1} \frac{(-1)^{2k}}{\sqrt{2k}+(-1)^{2k}}$$

I'm not sure where to go from here as far as simplifying goes. I tried adding the two fractions but I don't think that helps. Also, I can't use the Alternating series test on the original series because the absolute value of the terms is not monotone and decreasing.

2. Oct 19, 2009

### Dick

If k is an integer, then (-1)^(2k)=1 and (-1)^(2k-1)=(-1). Yes, you can't use the alternating series test. But you can learn something by combining the two terms and looking at the asymptotic form of the limit.

3. Oct 20, 2009

### utleysthrow

Would you mind clarifying what you said about the asymptotic form? I already tried adding the two terms but it looks a bit messy, and I'm lost as to what I should be looking for next.

$$\sum b_{k} = \sum \frac{-1}{\sqrt{2k-1}-1} + \sum \frac{1}{\sqrt{2k}+1} = \sum \frac{\sqrt{2k-1}-\sqrt{2k}-2}{\sqrt{4k^{2}-2k}+\sqrt{2k-1}-\sqrt{2k}-1}$$

Thank you.

4. Oct 21, 2009

### Dick

Can you show lim k->infinity sqrt(2k-1)-sqrt(2k)=0? If you ignore those terms then I would say that what's left behaves like -1/k for large k. That's what I meant by asymptotic form. That suggests that the way to prove it is to show |b_k|>C/k for some C for large k. Can you do that?