# Series Divergence - Apostol Calculus Vol I, Section 10.20 #7

Prove that $\displaystyle\sum_{n=2}^\infty\frac {(-1)^n}{\sqrt{n}+(-1)^n}$ diverges.

It is easy to see that this absolutely diverges, however how can it be proven to diverge in general? The idea as posted in another forum was to group the terms in the sequence as follows:

$$\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n}+(-1)^n}=-\sum_{n=1}^\infty \left(\frac{1}{\sqrt{2n+1}-1}-\frac 1 {\sqrt{2n}+1}\right)$$

and then show that this sequence diverges. I did end up getting an answer from that forum (thank you!) however it depended on results from WolframAlpha which seem extremely difficult to solve by hand. This question is from Apostol's Calculus, written in 1969, so I'd like to have a solution which doesn't depend on WolframAlpha.

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It is easy to see that this conditionally converges... Is it? Does this? – Did Oct 9 '11 at 21:38
Sure... doesn't it? $lim_{n\to\infty} \frac 1 {\sqrt{n}+(-1)^n}=0$ – process91 Oct 9 '11 at 21:41
So what? You might wish to state precisely the theorem you think you are applying and to check that every hypothesis of this theorem holds in your case. – Did Oct 9 '11 at 21:44
Ah, yes, that was the difficulty. It's been a while since I worked on this problem. The difficulty was proving that it did not conditionally converge. – process91 Oct 9 '11 at 21:46
Exactly. One term in the numerator, two in the denominator, each of them with a simple equivalent $cn^a$ for some $c$ and $a$... the way to the solution seems clear. – Did Oct 9 '11 at 21:55

$\rm HINT \text{ }$: $\text{ } \displaystyle \sum_{n=2}^\infty \left( \frac{(-1)^n }{\sqrt{n}} -\frac{(-1)^n}{\sqrt{n}+(-1)^n} \right)= \sum_{n=2}^{\infty} \frac{1}{n + (-1)^n \sqrt{n}} .$

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OK, so $\sum \frac {(-1)^n} {\sqrt{n}}$ converges, but $\sum \frac 1 {n+(-1)^n \sqrt{n}}$ diverges by the comparison theorem with $\sum \frac 1 {2n}$. This proves, by contradiction, that $\sum \frac {(-1)^n}{\sqrt{n}+(-1)^n}$ diverges. Correct? – process91 Oct 10 '11 at 3:38
@process91 Yes that's correct. – Ragib Zaman Oct 10 '11 at 4:17

How about $\left| \frac{(-1)^n}{\sqrt{n} + (-1)^n} \right| \ge \frac{1}{\sqrt{n}+1}$, and that $\sum_{n=2}^\infty \frac{1}{\sqrt{n}+1}$ diverges by integral test ?

Added: This answer the second version of the question, namely, to show that $\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n} + (-1)^n}$ diverges. As the OP already showed the sum can be rewritten as (notice that the lower bound is $1$ and not $2$ as in OP's question): $$\mathcal{S} = \sum_{n=1}^\infty \left(\frac{1}{\sqrt{2n} + 1 } - \frac{1}{\sqrt{2n+1}-1}\right)$$ Series expansion of the summand at infinity: $$\frac{1}{\sqrt{2n} + 1 } - \frac{1}{\sqrt{2n+1}-1} \sim -\frac{1}{n}+\frac{1}{4 \sqrt{2}} \frac{1}{n^{3/2}}+ O(n^{-2})$$ suggests that the series diverges as harmonic series. Indeed, note that $$\frac{1}{\sqrt{2n} + 1 } - \frac{1}{\sqrt{2n+1}-1}$$ is strictly increasing for $n\ge 1$, and thus integral test applies, which shows divergence.

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In what subject do you learn how to do series expansion at infinity? It seems like this is extremely useful, as the most common responses to the difficult series I have run into involve answers of this sort, however the book has not covered this yet. – process91 Oct 10 '11 at 7:08
@process91, had you pursued the algebra explained in the comments, the equivalent Sasha states above would have popped up without any series expansion. (But I have nothing against series expansions, mind you...) – Did Oct 10 '11 at 7:52
Yes, I figured that when I saw it, but I'm still unsure of how to proceed algebraically. I mean, it's not exactly equal to those results... I just haven't used that method. Could you give any further details on how to proceed that way? I worked it out to be $\frac{2+\sqrt{2n}-\sqrt{2n+1}}{\sqrt{2n^2+n}-\sqrt{2n}+\sqrt{2n+1}-1}$, but I don't see any other simplifications. – process91 Oct 10 '11 at 8:06
Keep the denominator written as a product of two terms. One term is $\sqrt{2n}+1\sim\sqrt2\sqrt{n}$, the other is $\sqrt{2n+1}-1\sim\sqrt2\sqrt{n}$ hence the denominator is $\sim2n$. In the numerator, $\sqrt{2n+1}-\sqrt{2n}\to0$ hence the numerator is $\sim2$. End of the proof. (I thought my previous comment explained exactly this.) – Did Oct 10 '11 at 10:42
It did, but the book has not really gone over these asymptotic approximations ~ in great detail yet, so it just didn't occur to me that this would be acceptable. Apostol mentions that $f(n) \sim g(n)$ means that $\lim_{n\to\infty} \frac {f(n)}{g(n)}=1$, however he doesn't really go over any examples which make use of that to prove that series converge, especially asymptotically approximating various parts separately like that, although obviously it makes sense. Thanks for posting the explanation! – process91 Oct 10 '11 at 15:48