# Checking the following series $\sum_{n=1}^{\infty}\frac{(-1)^n}{4\sqrt{n}}\frac{(16n-4)}{\sqrt{16n+64}}$

I'm stuck trying to check if the following series is convergent: $$\sum_{n=1}^{\infty}\frac{(-1)^n}{4\sqrt{n}}\frac{(16n-4)}{\sqrt{16n+64}}$$ I tried to use Leibniz‏ theorem but without any success. Is is possible to show how to prove it?

• well the numerator is $O(n)$ while the denominator is $O(\sqrt n) * O(\sqrt n) = O(n)$ so before you even start to check you can already know that the sequence behaves like $O(1)$ so the series must diverge. So instead of checking blindly you can use some method to try and prove that it's divergent (hint: since it's $O(1)$, it means that the sequence does not approach 0) – Francisco José Letterio Jan 31 at 15:11

## 2 Answers

A necessary condition for the convergence of a series $$\sum\limits_{n = 1}^\infty {a_n }$$ is that $$\mathop {\lim }\limits_{n \to + \infty } a_n = 0$$ Since $$\mathop {\lim }\limits_{n \to + \infty } \frac{1} {{4\sqrt n }}\frac{{16n - 4}} {{\sqrt {16n + 64} }} = 1$$ we have that $$\mathop {\lim }\limits_{n \to + \infty } a_n = \mathop {\lim }\limits_{n \to + \infty } \left( { - 1} \right)^n \frac{1} {{4\sqrt n }}\frac{{16n - 4}} {{\sqrt {16n + 64} }}$$ does not exists and the series is not convergent.

Hint: do the terms go to $$0$$?