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