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I am asked to determine whether the series $\sum_{n=1}^\infty \frac{2}{\sqrt{n}+2}$ converges or diverges.

I begin by looking if i can do a test for divergence but in this case i cannot tell anything about the divergence just by taking the limit $n \to \infty$. A direct comparison test seems like an option. One series of whose convergence/divergence i know is

$$2 \sum_{n=1}^\infty \frac{1}{\sqrt{n}}$$ this is a p-series with $p=\frac{1}{2}$ and it therefore diverges as $p \lt 1$. However $\frac{1}{\sqrt{n} +2} \lt \frac{1}{\sqrt{n}}$ and so this does not help. What else could i try?

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For $n>4$ we have $$\frac{1}{\sqrt n+ 2}>\frac{1}{n}$$ because for $n>4$ we have $n>\sqrt{n}+2$

Therefore as the series terms are is greater than the terms of the harmonic series which diverges, the given series diverges.

Hope this can be useful

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Note that $2 \sqrt n\ge 2$ and so

$$\frac{1}{\sqrt n+ 2}\ge \frac{1}{3\sqrt n}$$

Can you finish now?

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Note that $\sum_{n=1}^\infty \frac{2}{\sqrt{n}+2} $ contains a polynomial over polynomial expression and so the limit comparison test could be useful.

let $b_n = \frac{1}{\sqrt{n}}$ and $a_n = \frac{2}{\sqrt{n}+2}$

Taking $$\lim_{ n \to \infty} \frac{a_n}{b_n} = \lim_{ n \to \infty} \frac{2}{1+\frac{2}{\sqrt{n}}} = 2 \gt 1$$

This tells us that both $\sum a_n$ and $\sum b_n$ will converge or diverge dependent on the convergence or divergence of $\sum b_n$

now, $\sum_{n=1}^\infty b_n $ is divergent as it is a p-series with $p=\frac{1}{2}$ as you mentioned above and so the original series diverges.

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