How can one determine whether the following series converges or diverges

Question

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

Wolfram Alpha returns nothing useful, except that the ratio test was inconclusive.

Answer 1

One way of doing it, which is not necessarily the most sophisticated way to go about it, is to look at the partial sums and see if those trend toward some limit. Let's, for the sake of clear example, use the sum "$1 + 1/2 + 1/4 + 1/8 + ...$" if we add the first two terms, our partial sum is 1.5. For the first three it's 1.75. And for the first 4 it's 1.875. As you can seem the partial sums are trending toward two and if we had a function which gave us these partial sums it's limit as x approaches infinity would be two as well.

Answer 2

First, with any convergence problem, you should check that the sequence goes to 0 as n approaches infinity. In this case, it obviously does.

What you should do next is look at the denominator and realize that as n becomes very large, sqrt(n)>>(-1)^n. Thus, for much of the infinite series, you can approximate the denominator as sqrt(n). Then the argument is simply [(-1)^n]/[sqrt(n)]. As previously stated, the corresponding sequence goes to 0 as n approaches infinity. Furthermore, the numerator constantly changes sign with every value of n, while the denominator is always positive. Thus, by the alternating series test, this series converges.

Answer 3

If $$ a_n=\frac{(-1)^n}{(-1)^n+\sqrt{n}}, $$ then $$ a_{2n}+a_{2n+1}=\frac{1}{\sqrt{2n}+1}-\frac{1}{\sqrt{2n+1}-1}= \frac{\sqrt{2n+1}-\sqrt{2n}-2}{(\sqrt{2n+1}-1)(\sqrt{2n}+1)}=-\frac{1}{n}+{\mathcal O}(n^{-3/2}). $$ Hence, the series diverges to $-\infty$.