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

Question

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.

Answer 1

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 $$\sum_{n=1}^{80}\frac1{\sqrt n}<17-\frac19.$$

Answer 2

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 $b=80$

Answer 3

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