Question

$\underset{n\to \infty }{lim}\frac{\left[x\right]+\frac{1}{2}\left[2x\right]+\frac{1}{3}\left[3x\right]+...+\frac{1}{n}\left[nx\right]}{1^2+2^2+3^2+....+n^2}$
(where $[.]$ denotes the greatest integer)

• A. 0
• B. $\frac{1}{2}$
• C. $\frac{1}{6}$
• D. 1

Answer 1

The correct option is A 0

$\underset{n\to 0}{lim}\frac{\left[x\right]+\frac{1}{2}\left[2x\right]+\frac{1}{3}+.....+\frac{1}{n}\left[nx\right]}{1^2+2^2+3^2+.....n^2}$

$\underset{n\to 0}{lim}\frac{6\left[x\right][1+1+1+....n+n]}{n(n+1)(2x+1)}$

$\underset{n\to 0}{lim}\frac{6\left[x\right]\left[x\right]}{n(n+1)(2n+1)}=\underset{n\to 0}{lim}\frac{6\left[x\right]}{(n+1)(2x+1)}$

$\underset{n\to 0}{lim}\frac{6}{n^2}\frac{\left[x\right]}{(1+\frac{1}{n})(1+\frac{1}{n})}=\frac{1}{\infty }\times \frac{6\left[x\right]}{(1+0)(2+0)}$

$\Rightarrow \frac{0}{(1+0)(2+0)}\Rightarrow 0$

$\underset{n\to 0}{lim}\frac{\left[x\right]+\frac{2}{2}\left[2x\right]+.....+\frac{1}{x}\left[nx\right]}{1^2+2^2+3^2+....n^2}=0$