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Does the following series converge or diverge? $$ \sum\limits_{n=1}^\infty\frac{1}{\sqrt{n}+\sqrt{n+1}} $$ The methods I have at my disposal are geometric and harmonic series, comparison test, limit comparison test, and the ratio test.

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5 Answers 5

up vote 15 down vote accepted

It is not hard to see that $$\sum_{n=1}^\infty\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sum_{n=1}^\infty(\sqrt{n+1}-\sqrt{n})$$

As you know this series is divergent.

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We can easily see that $(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})=1$ –  Babak S. Jun 24 '12 at 15:40
By partial sums, correct? $\(\sqrt{2}-\sqrt{1}\)+\(\sqrt{3}-\sqrt{2}\)+...+\(\sqrt{n+1}-\sqrt{n}\)=-\sqrt{‌​1}+\sqrt{n+1}$ $\lim_{n\to\infty}-\sqrt{1}+\sqrt{n+1}=DNE$ –  user1405177 Jun 24 '12 at 15:54
@user1405177: What Babgen pointed here is exactly what you wanted. Easy approach. –  Babak S. Jun 24 '12 at 16:02

For $n\geq 1$, we have $\sqrt n+\sqrt{n+1}\leq 2\sqrt{n+1}\leq 2(n+1)\leq 4n$ hence $$\frac 1{\sqrt n+\sqrt{n+1}}\geq \frac 1{4n}\geq 0$$ and we can conclude using the fact that the harmonic series $\sum_{k=1}^{+\infty}\frac 1k$ is divergent.

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Thank you very much! –  user1405177 Jun 24 '12 at 15:32

You also have all of your experience with limits and approximation available. The key observation that makes things 'obvious' is that $\sqrt{n+1} \approx \sqrt{n}$, and so

$$ \frac{1}{\sqrt{n} + \sqrt{n+1}} \approx \frac{1}{2\sqrt{n}} $$

and so you can apply your knowledge about the convergence of sums of the form $\sum 1/n^s$. For example, since $s = 1/2$, this should diverge faster than the harmonic series - a lot faster really - and so you should have no trouble comparing the original sum to the harmonic series (e.g. as in Davide's answer).

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You may use the simple fact that $n\ge \sqrt n$ when $n\ge 1$.Then by using a trivial inequality we get that:

$$\sum\limits_{n=1}^\infty\frac{1}{2(n+1)}=\frac{1}{2}(H_n-1) \rightarrow \infty\le\sum\limits_{n=1}^\infty\frac{1}{n + n+1}\le\sum\limits_{n=1}^\infty\frac{1}{\sqrt{n}+\sqrt{n+1}}$$


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You can also check out this step by step explanation: http://www.symbolab.com/solutions?query=%5Csum%20_%7Bn%3D1%7D%5E%7B%5Cinfty%20%7D%5Cfrac%7B1%7D%7B%5Csqrt%7Bn%7D%2B%5Csqrt%7Bn%2B1%7D%7D

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This should be a comment, not an answer. –  Cameron Williams Sep 9 '13 at 14:53
@Cameron, I would appreciate if you can explain. –  John Sep 9 '13 at 14:58
I just realized that you're a new member and probably don't have enough reputation to comment. Generally, the community prefers answers to be contained within the answer itself (and not linked to outside pages). You should edit your answer so I can remove the downvote. –  Cameron Williams Sep 9 '13 at 15:03
Answers on math.SE should be as self-contained as possible. This answer consists of little more than a link to an external site, meaning that if that site were to ever disappear (or even just change its link URLs) your answer would be entirely meaningless. Please STOP posting link-only answers, and begin including more of the pertinent information in your posts (with appropriately cited sources, of course). –  Arthur Fischer Sep 9 '13 at 18:55

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