Prove 16<1/√(1)+1/√(2)+1/√(3)+1/√(4)+...1/√(79)+1/√(80)< 17

Show that [MATH]16 < \sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}< 17[/MATH].

It goes without saying that there's more than one way to skin any given math problem. Exposing students with intriguing math problem can actually instill a genuine interest in creative problem-solving from an early age.

Educators must bear in mind that students usually get quite bored with mathematics when they are asked to tons of similar problems. Therefore, it is entirely up to the educators, math teachers to shower their students the good and challenging problems so students would find math a fun and exciting subject to work with.

This really boils down to the ability of the educators to select good problems as exercises/contest math problems. Math educators should have a large collection of challenging math problems at hand to train them to become a better and competent problem solvers.

This particular problem fits the category of intriguing and good math problem. That is why it earns a place at this blog.

I will show you my solution how to tackle this problem partially, i.e. to prove the upper bound for this problem, [MATH]16<\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}[/MATH].

We could use trapezoid rule to estimate [MATH]\int_{1}^{81} \dfrac{1}{\sqrt{x}}\,dx[/MATH] and since the function $\dfrac{1}{\sqrt{x}}$ is concave up, the result obtained from the trapezoid rule will be larger than the integral.

Therefore we have:

[MATH]\int_{1}^{81} \dfrac{1}{\sqrt{x}}\,dx < \dfrac{1}{2}(1)\left(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}\right)+\dfrac{1}{2}(1)\left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}\right)+\cdots+\dfrac{1}{2}(1)\left(\dfrac{1}{\sqrt{79}}+\dfrac{1}{\sqrt{80}}\right)+\dfrac{1}{2}(1)\left(\dfrac{1}{\sqrt{1}}+\dfrac{80}{\sqrt{81}}\right)[/MATH]

[MATH]\left[2x^{\frac{1}{2}}\right]_{1}^{81} < \dfrac{1}{2}+\left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\cdots+\dfrac{1}{\sqrt{80}}\right)+\dfrac{1}{18}[/MATH]


[MATH][2(9)-2(1)]<\left(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\cdots+\dfrac{1}{\sqrt{80}}\right)+\dfrac{1}{18}-\dfrac{1}{2}[/MATH]

[MATH]16<16+\dfrac{4}{9}<\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\cdots+\dfrac{1}{\sqrt{80}}[/MATH]

Therefore we can say we have proved the upper bound for the given inequality.

But the thing is, can we prove the lower bound using the similar attack?