# Algebra Precalculus – Proving the Sum of Series with Floor Function

algebra-precalculusceiling-and-floor-functionsinequality

Consider this $$\left[\dfrac 12\displaystyle\sum_{n=1}^{k^2}\frac 1{\sqrt n}\right]$$ where $$[\cdot]$$ is the greatest integer function.

I had observed that its value is $$(k-1)$$ by putting $$k=2,3,4,$$ upto $$5$$ and pressing my calculator button.

But could not prove it manually.

I thought about this inequalities like $$n>\sqrt n\implies \frac 1{\sqrt n}>\frac 1 n$$ but got nothing. Some clever approach is needed here Which I am unable to find.

As $$f(x)=\dfrac{1}{2\sqrt{x}}$$ is strictly decreasing, for every integer $$n$$ we have $$f(n)=\displaystyle\int_{n}^{n+1}f(n)\,dx> \displaystyle\int_{n}^{n+1}f(x)\,dx$$ and $$f(n)=\displaystyle\int_{n}^{n+1}f(n)\,dx< \displaystyle\int_{n-1}^{n}f(x)\,dx$$ It follows from the first inequality that $$\displaystyle\sum_{n=1}^{k^2}f(n)> \displaystyle\int_{1}^{k^2+1} f(x)\,dx=\displaystyle\int_{1}^{k^2+1}\dfrac{1}{2\sqrt{x}}dx=\sqrt{k^2+1}-1$$
As $$k^2+1\geq k^2$$, we have $$\displaystyle\sum_{n=1}^{k^2}f(n)> \sqrt{k^2}-1=k-1$$.
And from the second inequality, $$\displaystyle\sum_{n=1}^{k^2}f(n)< \displaystyle\int_{0}^{k^2} f(x)\,dx=\displaystyle\int_{0}^{k^2}\dfrac{1}{2\sqrt{x}}dx=\sqrt{k^2}=k$$
This proves that $$\left\lfloor \displaystyle\sum_{n=1}^{k^2}\dfrac{1}{2\sqrt{n}}\right\rfloor=k-1,$$ as desired.