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

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

On my previous post (Show that [MATH]16<\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{80}}<17[/MATH]), we used the trapezoid rule to prove the upper bound for the given inequality, i.e. [MATH]16<\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}[/MATH]. We're then not supposed to use the same method to prove the lower bound simply because the function $y=\dfrac{1}{\sqrt{x}}$ is concave up. No matter how we manipulated that concept, we will only end up with proving the target sum [MATH]\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}[/MATH] will be greater than some quantity, not less than. This works against to what we are looking to prove, that is, [MATH]\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}<17[/MATH].