Proving divergence of $\sum_2^\infty{(-1)^n}{1\over \sqrt n+(-1)^n}$ Q: Is $\sum_2^\infty{(-1)^n}{a_n}$ convergent?
$$a_n={1\over \sqrt n+(-1)^n} \space \forall n\in\mathbb{N_{\ge2}}$$
The answer says it diverges.The only thing I could deduce is that the series is not absolutely convergent by comparison test.I  tried to find a smaller diverging series but failed miserably.I guess only thing that can be done is to show the series doesn't meet
the Cauchy criterion but the difference of two partial sums looks horribly refractory to me.
 A: We have
$${(-1)^n\over \sqrt n+(-1)^n}=\frac{(-1)^n}{\sqrt n}\frac{1}{1+\frac{(-1)^n}{\sqrt n}}=\frac{(-1)^n}{\sqrt n}\left(1-\frac{(-1)^n}{\sqrt n}+O\left(\frac 1 n\right)\right)\\=\frac{(-1)^n}{\sqrt n}-\frac{1}{n}+O\left(\frac 1 {n^{3/2}}\right)$$
so the given series is sum of three series:


*

*the first is convergent by Leibniz rule

*the second is the harmonic divergent series

*the third is convergent by comparison with a Riemann series. Conclude.

A: Just group the terms in pairs and look for a way to estimate $b_n = a_{2n} - a_{2n+1}$. If the sum of $b_n$ diverges then so does the original (actually the two series are equivalent because $a_n\to 0$).
A: \begin{align}
a_{2n}+a_{2n+1}&=\frac{1}{\sqrt{2n}+1}-\frac{1}{\sqrt{2n+1}-1}=\frac{-2+(\sqrt{2n+1}-\sqrt{2n})}{\sqrt{2n(2n+1)}+\sqrt{2n+1}-\sqrt{2n}-1} \\ &=
\frac{-2-\frac{1}{\sqrt{2n}+\sqrt{2n+1}}}{2n\sqrt{1+\frac{1}{2n+1}}-\frac{1}{\sqrt{2n+1}+\sqrt{2n}}-1}=-\frac{1}{n}\left(\frac{1+\frac{1}{2\sqrt{2n}+2\sqrt{2n+1}}}{\sqrt{1+\frac{1}{2n+1}}-\frac{1}{2n\sqrt{2n+1}+2n\sqrt{2n}}-\frac{1}{2n}}\right).
\end{align}
Clearly
$$
n(a_{2n}+a_{2n+1})\to -1.
$$
In fact
$$
\left|\,a_{2n}+a_{2n+1}+\frac{1}{n}\right|\le \frac{k}{n^{3/2}},
$$
for a suitable $k>0$
and hence the series it diverges.
