# Alternate series [duplicate]

The alternate series $S=\displaystyle \sum_{k=2}^{\infty} \frac{(-1)^n}{\sqrt{n}+(-1)^n}$ converges?

$S$ is absolutely convergent?

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## marked as duplicate by David Mitra, Martin Sleziak, Daniel Fischer♦, Tomás, GinaJul 21 '14 at 14:30

sorry to ask this, but what is the idea behind converging? I am hearing a lot on this site.. – MonK Jul 21 '14 at 13:31
have you done any work here? – 5xum Jul 21 '14 at 13:31
@Sid Converging is when a series, function, etc. gets closer to a specific value. Like how 0.9, 0.99, 0.999... gets closer to 1. – Eul Can Jul 21 '14 at 13:33
y tried using leibniz rule but not works $a_n=\frac{1}{\sqrt(n)+(-1)^n}$ is not a monotonic sequence and not converges to 0 and I don't know calculate $\lim an$ – Pablo Herrera Jul 21 '14 at 13:37
So in this question, S is getting or will get closer to 1? (a specific arbitrary value) – MonK Jul 21 '14 at 13:38

We have using Taylor series

$$\frac{(-1)^n}{\sqrt{n}+(-1)^n}=\frac{(-1)^n}{\sqrt n}\left(1+\frac{(-1)^n}{\sqrt n}\right)^{-1}=\frac{(-1)^n}{\sqrt n}-\frac1n+\mathcal O\left(\frac1{n^{3/2}}\right)$$ so we see that the given series is the sum of $3$ series: two convergent series using Leibniz theorem and using the comparison with a Riemann series and a harmonic series which is divergent. Hence the given series is divergent.

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taylor series seriously? it never be in my plans hahaha thanks a lot – Pablo Herrera Jul 21 '14 at 13:44
You're welcome. – user63181 Jul 21 '14 at 13:45