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Does the series $$ \sum_{n=1}^{\infty}\frac{\sqrt{n+2}-\sqrt{n}}{n^{3/2}} $$ converge or diverge?

My attempt was to write the series as $$ \sum_{n=1}^{\infty}\frac{\sqrt{n+2}-\sqrt{n}}{n^{3/2}}=\sum_{n=1}^{\infty}\frac{\sqrt{n+2}}{n^{3/2}}-\sum_{n=1}^{\infty}\frac{1}{n} $$

The first series can be estimated from below by the harmonic series: $$ \sum_{n=1}^{\infty}\frac{\sqrt{n+2}}{n^{3/2}}=\sum_{n=1}^{\infty}\frac{(n+2)^{1/2}}{n^{1/2}\cdot n}\geqslant\sum_{n=1}^\infty \frac{1}{n}=\infty $$ and hence diverges.

The second series is the harmonic series and hence diverges.

Now, I am not sure what the whole thing does.

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Your attempt is wrong because you can't separate the series into a difference of divergent series. You got an indeterminate form $\infty - \infty$.

You could try to rationalize the numerator: $$(\sqrt{n+2}-\sqrt{n})\frac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}+\sqrt{n}}=\frac{2}{\sqrt{n+2}+\sqrt{n}}$$ Then the series becomes

$$\sum_{n=1}^{\infty}\frac{2}{(\sqrt{n+2}+\sqrt{n})n^{3/2}}$$

You can use comparison test to conclude.

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  • $\begingroup$ I suggest to do not give full answer suddenly when possible, since it can be not useful for the site nor for the asker who could be interested to solve it by themself. In such latter cases the use of "spoiler" comand can be useful). $\endgroup$ – user Dec 8 '18 at 13:21
  • $\begingroup$ Using "spoiler" command is a good idea $\endgroup$ – Lorenzo B. Dec 8 '18 at 13:23
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    $\begingroup$ This wasn't and isn't a full answer. It leaves the determination of convergence/divergence to the asker. It merely suggests rationalizing the numerator, and displays the result upon doing so. I think it is a fine answer. Nice work, @LorenzoB. $\endgroup$ – amWhy Dec 8 '18 at 13:30
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    $\begingroup$ @Rhjg, yes you are $\endgroup$ – Lorenzo B. Dec 8 '18 at 13:38
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    $\begingroup$ Thank you very much for your help. $\endgroup$ – Rhjg Dec 8 '18 at 13:39
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HINT

We can more effectively use that

$$\sqrt{n+2}-\sqrt{n}=\frac{2}{\sqrt{n+2}+\sqrt{n}}$$

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You can use the estimate: $$\frac{\sqrt{n+2}-\sqrt{n}}{n^{3/2}}<\frac1{n^2} \iff \sqrt{n}(\sqrt{n+2}-\sqrt{n})<1 \iff \\ \sqrt{n(n+2)}<n+1 \iff n^2+2n<n^2+2n+1.$$

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Rewriting :

$2=(n+2) -n=$

$(\sqrt{n+2}-√n)(\sqrt{n+2}+√n) \gt$

$\sqrt{n+2}-√n$, since

$\sqrt{n+2}+√n >1$.

Hence

$\dfrac{\sqrt{n+2}-√n}{n^{3/2}} \lt \dfrac{2}{n^{3/2}}.$

Use comparison test.

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Note that $\sqrt{n+2}-\sqrt n\to 0$ as $n\to\infty$, so $\;\dfrac{\sqrt{n+2}-\sqrt{n}}{n^{3/2}}=o\biggl(\dfrac 1{n^{3/2}}\biggr)$, and the latter series converges.

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  • $\begingroup$ That's a nice way but why $\sqrt{n+2}-\sqrt n\to 0$? $\endgroup$ – user Dec 8 '18 at 13:37
  • $\begingroup$ That is standard in high school (and your answer shows how it is proved). $\endgroup$ – Bernard Dec 8 '18 at 13:42
  • $\begingroup$ Yes but it seems that the asker ignore that otherwise probably he wouldn't have posted the question. In my opinion at the present state you answer is not complete regarding a crucial point. Moreover once we know it goes to zero we could simply use comparison test. Please consider my observations in a constructive way. $\endgroup$ – user Dec 8 '18 at 13:46
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    $\begingroup$ @gimusi : I think our answers are more or less complementary. They do not aim the same audience, and I wanted to propose an as concise as possible answer for those who already have some familiarity with the basics of limits. $\endgroup$ – Bernard Dec 8 '18 at 13:57
  • $\begingroup$ That's ansolutely fine my aim was only to suggest what I thought could make some improvements with that for the asker. But that's of corse all up to you and the answer is fine also in that way. Bye $\endgroup$ – user Dec 8 '18 at 14:08

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