Convergence of ∑(−1)^n/(√n+(−1)^n) using paired terms

[utleysthrow]

Homework Statement

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

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.

Homework Equations

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.

 

[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.

 

[utleysthrow]

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.

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.

 

[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?