# How can I prove that $16 \lt {1+\frac1{\sqrt2}+\frac1{\sqrt3}+\cdots+\frac1{\sqrt{80}}<18}$?

What i want to prove is this $$16 \lt {1+\frac1{\sqrt2}+\frac1{\sqrt3}+\cdots+\frac1{\sqrt{80}}<18}$$

I haven't encountered any problem of this kind before, how do we proceed?

Making approximations dosen't seem feasible, so all the suggestions are welcome.

• Welcome to Mathematics Stack Exchange. Did you mean $< 18$? Commented Jul 9, 2020 at 18:32
• @J.W.Tanner, yes, that was a typo Commented Jul 9, 2020 at 18:36
• If you haven't been introduced to calculus yet, then why do you need to solve this exercise? Commented Jul 9, 2020 at 18:39
• I think the idea is in the following. Try to show by induction, that, starting from some $n$, $2(\sqrt{n}-1) <1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}< 2 \sqrt{n}$. It should be something like that. However I'm not sure, that why i'm writing it as a comment. Commented Jul 9, 2020 at 18:46
• There are interesting ideas in this post. math.stackexchange.com/q/2149448/399263 In particular you can use $\sum \frac 1{\sqrt{k}}>2(\sqrt{n+1}-1)=16$ for $n=80$ using telescoping sum.
– zwim
Commented Jul 9, 2020 at 18:47

if we have $$g(x) > 0$$ but $$g'(x) < 0,$$ then $$\int_a^{b+1} \; g(x) \; dx \; < \; \sum_{j=a}^b \; g(j) \; < \; \int_{a-1}^b \; g(x) \; dx$$

for you $$g(x) = \frac{1}{\sqrt x}$$ which is integrable at the origin, so we can take $$a=1$$ and $$b=80$$

• The OP does not know this: $\displaystyle\int$...
– user65203
Commented Jul 9, 2020 at 18:47
• @YvesDaoust you seem to be correct. I like Dietrich's question. Commented Jul 9, 2020 at 18:49

Let's look at the telescoping sum $$S=(\sqrt2-1)+(\sqrt3-\sqrt2)+(\sqrt4-\sqrt3)+\cdots+(\sqrt{81}-\sqrt{80})=9-1=8.$$ Then $$S=\sum_{n=1}^{80}(\sqrt{n+1}-\sqrt n)=\sum_{n=1}^{80}\frac1{\sqrt{n+1}+\sqrt n}.$$ So $$S<\sum_{n=1}^{80}\frac1{2\sqrt n}$$ so $$\sum_{n=1}^{80}\frac1{\sqrt n}>2S=16.$$ Also $$S>\sum_{n=1}^{80}\frac1{2\sqrt{n+1}}$$ so that $$\sum_{n=1}^{80}\frac1{\sqrt{n+1}}<2S=16.$$ But $$\sum_{n=1}^{80}\frac1{\sqrt n} =\sum_{n=1}^{80}\frac1{\sqrt{n+1}}+1-\frac19$$ and we get $$\sum_{n=1}^{80}\frac1{\sqrt n}<17-\frac19.$$

• More generally, $$2(\sqrt{b+1}-\sqrt a)\le\sum_{n=a}^b\frac1{\sqrt{n}}\le2(\sqrt{b+1}-\sqrt a)+\frac1{\sqrt a}-\frac1{\sqrt{b+1}}.$$
– user65203
Commented Jul 9, 2020 at 19:04

This can be done using a theorem

$$2(\sqrt{n+1}-{n}) \lt \frac1 {\sqrt n}\lt 2(\sqrt n -\sqrt {n-1 })$$

And to prove it, just rationalise the numerator on RHS and LHS