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I know intuitively why this series diverges but I can't really get a proof.

So I am trying to use that the fact that: $\sum\limits_{n=1}^\infty\frac{1}{2+\sqrt{n}} > \sum\limits_{n=1}^\infty\frac{1}{2+{n}} $.

And then from there I want to get it down to something like 1/n, which is a p-series with p=1 therefore it converges, and then use the comparison test to show the original sequence diverges. However I am stuck finding in this middle bit.

Any help would be appreciated.

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  • $\begingroup$ Limit comparison? $\endgroup$ – Angina Seng May 30 '19 at 5:29
  • $\begingroup$ How can you say the last one (with the -3/2) diverges? I understand why it would. But are you able to just say it does without anything further? $\endgroup$ – user11015000 May 30 '19 at 5:36
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You already have the answer. $\sum_{n=1}^\infty\frac1{2+n}$ is a tail of the divergent harmonic series and thus diverges itself. The original series therefore diverges by the limit comparison test.

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  • $\begingroup$ Could you explain more when you say "tail of the divergent harmonic series"? Thanks. $\endgroup$ – user11015000 May 30 '19 at 5:35
  • $\begingroup$ @user11015000 A tail of a series is everything but the first $k$ terms, where $k$ is finite. Since the first $k$ terms must sum to a finite number, whether such a tail diverges or not is the same as whether the entire series diverges or not. $\endgroup$ – Parcly Taxel May 30 '19 at 5:39
  • $\begingroup$ @user11015000 You can see that by tail one means terms as $n \to \infty$. Since as $\lim\limits_{n \to \infty} \frac{1}{2+n}$, follows similar behaviour as $\lim\limits_{n \to \infty}\frac{1}{n}$, the tail diverges. $\endgroup$ – Rick May 30 '19 at 5:42
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    $\begingroup$ @user11015000 $\sum_{n=1}^\infty \frac{1}{2+n} = \sum_{n=3}^\infty \frac{1}{n}$ $\endgroup$ – jawheele May 30 '19 at 5:56
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    $\begingroup$ @user11015000 Suppose this tail converges. Then the entire series converges, for the head must converge, but it doesn't - contradiction. $\endgroup$ – Parcly Taxel May 30 '19 at 6:01

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