Algebra Precalculus – Proving the Sum of Series with Floor Function


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.

Best Answer

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.