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I'm trying to figure out the limit of the sequence

\begin{equation*} \lbrace\sum_{n=1}^{k}\frac{1}{\sqrt{k^2 + n}}\rbrace_{k=1}^\infty \end{equation*}

I feel like I'm close to understanding it, I just need a little push. I understand that I need to show that this sequence lies between $\lbrace \frac{k}{\sqrt{k^2 + k}}\rbrace_{k=1}^\infty$ and $\lbrace \frac{k}{\sqrt{k^2 + 1}}\rbrace_{k=1}^\infty$, and that both of those sequences converge to 1. However, I'm not quite sure how to show that both of them converge to 1, or why the first sequence would lie between them. The summation and the square roots are confusing me, I think.

Edit: Okay, so I have the notion that I should divide everything in the convergent sequences by k, but I don't think I'm doing the algebra right. For the first sequence, dividing by k seems to get me $\frac{1}{\sqrt{k + 1}}$, which seems to converge to 0 as k approaches $\infty$ since $\frac{1}{\sqrt{k + 1}} \leq \frac{1}{k}$. Similarly, dividing everything in the latter sequence seems to give $\frac{1}{\sqrt{k + \frac{1}{k}}}$, which would also converge to zero as k approaches $\infty$. I don't feel like I'm distributing the k properly over the square root.

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Is this homework? – Aryabhata Sep 28 '10 at 1:40
Yes, sorry. I see from looking at other similar questions on here that I should tag it as such. Thanks. – Jarrod Sep 28 '10 at 1:58
up vote 3 down vote accepted

Since this looks like homework, I will give a hint.

Which is the largest among $\sqrt{k^2+1}, \sqrt{k^2+2},\dots, \sqrt{k^2+k}$?

For your next step:

I think the mistake you are doing is substituting

$\frac{\sqrt{k^2+k}}{k}$ with $\sqrt{\frac{k^2+k}{k}}$ instead of $\sqrt{\frac{k^2+k}{k^2}}$.

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Ah, so I kind of see that the original sequence as k goes to $\infty$ would be k terms of the form $\frac{1}{\sqrt{k^2 + n}}$. So then $\frac{k}{\sqrt{k^2 + k}}$ would be smaller than the smallest term and $\frac{k}{\sqrt{k^2 + 1}}$ would be larger than the largest term for all k. I'm still not sure how they converge to 1. I feel like there's a trick I'm missing. – Jarrod Sep 28 '10 at 1:54
Right. Now in those two terms, can you rewrite it in the form $\frac{1}{\sqrt{something}}$? – Aryabhata Sep 28 '10 at 2:31
As I said in response to yrudoy's answer, I tried dividing everything by k already but I think I'm making some sort of silly algebra error because it didn't come out right. Is there something special I need to do to distribute across a square root? – Jarrod Sep 28 '10 at 2:50
@Jarrod: Perhaps if you edit your question to show what you have so far, we can help further. – Aryabhata Sep 28 '10 at 3:10
Thank you for the suggestion. Done. – Jarrod Sep 28 '10 at 3:33

You want to take the limits of your two sequences as they approach infinity. if you are in a calculus class, you can use L'Hôpital's rule ( Otherwize, you should devide both the numerator and denominator by k, and then bring the k in the denominator into the radical. Then, as you go to infinity the stuff inside the radical goes to 1, and since 1/sqrt(1)=1, that becomes your answer.

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Is there something I need to do to distribute the k in the denominator inside the radical? I understand what you're saying but when I tried that earlier, it didn't come out right. When I distribute the k over the radical, does it become $\sqrt{k}$? – Jarrod Sep 28 '10 at 2:38

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