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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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).

Answer 4

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.

Answer 5

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.

Answer 6

Since, $\sqrt{n}\le \sqrt{n+1}\le n+1$ we have, $$\sum\limits_{n=1}^\infty\frac{1}{\sqrt{n}+\sqrt{n+1}} \ge \sum\limits_{n=1}^\infty\frac{1}{2\sqrt{n+1}} \ge \sum\limits_{n=1}^\infty\frac{1}{2(n+1)} =\infty$$

Answer 7

clearly the series is divergent but it can be assigned an analytical continuation sum, it could be demonstrated using $$\sum _{j=1}^{\infty } \left(\frac{1}{\sqrt{j}+\sqrt{j+1}}-\frac{\sqrt{\frac{1}{j}}}{2}\right)+\frac{\zeta \left(\frac{1}{2}\right)}{2}=-1$$ and $$\sum _{j=0}^{\infty } -\left(\sqrt{j+2}-\sqrt{j+3}\right)=-\sqrt{2}$$ Surely they will vote negatively, they will close it, but well, there is the simple demonstration