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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?

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  • $\begingroup$ 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) $\endgroup$ – Francisco José Letterio Jan 31 at 15:11
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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.

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Hint: do the terms go to $0$?

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