# Does $\sum\limits_{n=1}^\infty\frac{1}{\sqrt{n}+\sqrt{n+1}}$ converge?

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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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
I would prove the RHS is divergent by converting it to the LHS... – djechlin Nov 11 '15 at 17:52

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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It was not entirely obvious to me that the infinite sum of differences between square roots diverges, so I did telescoping:

$$\sum_{n=1}^{\infty} \left(\frac{1}{\sqrt{n+1} + \sqrt{n}}\right)=\sum_{n=1}^{\infty} \left(\frac{\sqrt{n+1} - \sqrt{n}}{n+1-n}\right)=\sum_{n=1}^{\infty} \left(\sqrt{n+1} + \sqrt{n}\right)=\\ =\lim_{N\to\infty} \sum_{n=1}^{N} \left(\sqrt{n+1}-\sqrt{n}\right)=\lim_{N\to\infty}\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\dots +\sqrt{N+1}-\sqrt{N}\right)=\\=\lim_{N\to\infty} \left(\sqrt{N+1}-\sqrt{1}\right)$$

So the limit of partial sum is equal to infinity.

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