# 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.

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.

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.

• We can easily see that $(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})=1$ Jun 24, 2012 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}$ $\lim_{n\to\infty}-\sqrt{1}+\sqrt{n+1}=DNE$ Jun 24, 2012 at 15:54
• @user1405177: What Babgen pointed here is exactly what you wanted. Easy approach. Jun 24, 2012 at 16:02
• I would prove the RHS is divergent by converting it to the LHS... Nov 11, 2015 at 17:52

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).

• All the answers on this question seem fine, but this one seems to me to be the best because it imparts intuition that will be useful in wider classes of problems. (I think the biggest failing of analysis, as a discipline, is the emphasis on writing perfectly-packaged proofs rather than conveying intuition.) Nov 21, 2017 at 16:18

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.

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.

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

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