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I am trying to prove that

$$\sum_{n=1}^\infty \frac{n}{\sqrt{n+1}}$$

diverges without checking the limit, bounds or doing any other lengthy steps, as it should be seen as divergent "immediately", but I have no clue about how I would quickly prove this.

So far I thought about using the P-series convergence test where it only converges for $p>1$ but it does not seem to make any sense for this one. I also thought about comparing it to other series but nothing comes to my mind.

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Use the term test: the $n$th term doesn't approach $0$ as $n\to\infty$.

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  • $\begingroup$ Thank you, I can't believe I forgot about using this! $\endgroup$ – Lightsong Dec 15 '19 at 12:17
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The sequence $a_n = \frac{n}{\sqrt{n+1}}$ is asymptotic to $b_n ={\sqrt{n+1}}$: $$ \lim_{n\to \infty} \frac{a_n}{b_n} = \lim_{n\to \infty} \frac{n}{n+1} = 1 $$ What can you say about the following series? $$\sum_{n=1}^{\infty}\sqrt{n+1}$$

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  • $\begingroup$ Thanks! I was also able to prove it this way too, since that series diverges with the sum going to infinite, and it has the same character as the original one, both are divergent $\endgroup$ – Lightsong Dec 15 '19 at 12:20
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Use that $$\frac{n}{\sqrt{n+1}}\geq \frac{1}{n}$$ and this is equivalent to $$n^4-n-1\geq 0$$ this is true for $$n\geq 2$$

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  • $\begingroup$ I guess this would fall into the comparison test? But to use this example, I would need to argue that the first term(s) are not exactly relevant in the convergence, right? $\endgroup$ – Lightsong Dec 15 '19 at 12:38
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You can notice that this series has the same behavior of $$\sum_{n=1}^\infty \frac{n}{\sqrt n} =\sum_{n=1}^\infty \sqrt n$$ that diverges.

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$$n+1 \le 2n\ \ \forall \ n \ge 1$$ $$\Rightarrow \frac{n}{\sqrt{2n}} \le \frac{n}{\sqrt{n+1}}$$ $$\Rightarrow \sum_{n=1}^{\infty}\frac{n}{\sqrt{2n}} \le \sum_{n=1}^{\infty}\frac{n}{\sqrt{n+1}}$$ $$\Rightarrow \frac{1}{\sqrt{2}}\sum_{n=1}^{\infty}{\sqrt{n}} \le \sum_{n=1}^{\infty}\frac{n}{\sqrt{n+1}}$$ Hence this won't converge!

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