# Proving $\sum_{n=1}^{99}\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}\lt\frac9{20}$

I found the original question asked by someone else, asking for this to be proven using only '9th grade math', this is the image:

Which can be written like

$$\sum_{n=1}^{99}\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}\lt\frac9{20}$$ Rationalizing it, I got

$$\sum_{n=1}^{99}\frac{1}{(2n+1)(\sqrt{n+1}+\sqrt{n})}<\frac9{20}$$

This is where I'm stuck. My plan was to somehow turn this into a telescoping sum, but I had no luck with that because of how messy the denominator can get in partial sums. I've tried to see if it followed the pattern of a GM series but the resulting ratio had radicals in. I generally don't know how to do sums of radicals. So how can I prove this inequality, preferably without using induction or manually evaluating each term using a calculator (because of the 9th grade math condition)? Any help is appreciated.

• The inequality is fairly sharp: $0.4395... < 0.45$. Commented Nov 8, 2021 at 14:36
• An easy bound would be $\frac 13 \sum_{n=1}^{99} (\sqrt{n+1}-\sqrt{n}) = \frac{9}{3}$. The one you propose is way sharper. Commented Nov 8, 2021 at 14:37
• Where did you find the original question? Commented Nov 8, 2021 at 14:47
• It is from a fellow student on a STEM Discord server. I can redirect any queries to them if need be.
– Typo
Commented Nov 8, 2021 at 14:49
• @MartinR and the infinite sum $\sum_{n=1}^\infty \frac{\sqrt{n+1} - \sqrt{n}}{2n+1} \approx 0.4895 > \frac{9}{20}$, so the upper limit $99$ has to figure into the answer somehow. Commented Nov 8, 2021 at 14:50

As we know $$(2n+1)^2=4n^2+4n+1>4n^2+4n=4n(n+1)$$, then $$2n+1>2\sqrt{n(n+1)}$$, with $$n\in\mathbb N$$. Hence we have $$\sum_{n=1}^{99}\frac{\sqrt{n+1}-\sqrt{n}}{\color{red}{2n+1}}<\sum_{n=1}^{99}\frac{\sqrt{n+1}-\sqrt{n}}{\color{red}{2\sqrt{n(n+1)}}}=\frac{1}{2}\sum_{n=1}^{99}\frac{1}{\sqrt n}-\frac{1}{\sqrt{n+1}}=\frac{1}{2}\left(1-\frac{1}{10}\right)=\frac{9}{20}.$$