- #1

arhzz

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**Summary::**I am suspossed to find the limit of this series.I've come to realize that the series diverges and I'm trying to prove that using the a comparison test.

Hello!

Consider this sum

$$ \sum_{k=1}^{n} (\sqrt{1+k} - \sqrt{k}) $$ the question wants me to find the limit of this sum where n is approching infinity.My first hunch was that this series is diverging and I wanted to prove that using the comparison test.I first tried to find a diffrent series that diverges and that it is smaller than my original series so (an>bn) and prove it that way.So what I did is I expanded the denominTOR (I am not sure if that is the right term in English but I think you will get what I meant)

$$

\sqrt{k+1}-\sqrt{k}=\frac{\left(\sqrt{k+1}-\sqrt{k}\right)\cdot\left(\sqrt{k+1}+\sqrt{k}\right)}{\sqrt{k+1}+\sqrt{k}}=\frac{1}{\sqrt{k+1}+\sqrt{k}}

$$

Now I have a series that is smaller than my original one,so now I need to prove that this series diverges.Since this isn't a "standard series" (harmonic,telescopic...) I cannot directly see if it diverges.So I tried "modifying" this series to one of the series in which I can easily determine if it diverges or not and go on from there.

What I did here is I've transformed the roots of the denominator like this.

$$\frac{1}{(k+1)^{1/2} + k^{1/2}} $$ Now this should be a harmonic series ## \frac{1}{k^s} ## where S is the exponent.Now if the exponent is greater than 1 the series converges,if it is smaller it diverges.Hence this series is divergent,which means the original series is also divergent.According to an online calculator this is correct,but I am not sure if the methods I used are "allowed" (if the math adds up).Would anyone be able to confirm if this is legit or not? Also if not what do I need to redo?

PS: I've tried looking in the "Homework help" department to post my question (since it is homework related) but I only found forums regarding physics and engineering.If this question doesn't belong here,please let me know in which forum I should post it.

Thank you and excuse the long post!

**[Moderator's note: moved from a technical forum.]**