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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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4 Answers 4

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:

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

Q.E.D.

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