Question

If the value of $\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n (\lfloor \frac{2n}{k} \rfloor - 2\lfloor \frac{n}{k}\rfloor)$ can be expressed as $(\log a - b)$ where a and b are positive integers, then $\frac{(a+b)}{4}$ is (where [.] denotes greatest integer function)

• A. 5/4
• B. 3/2
• C. 7/4
• D. 1

Answer 1

Answer: (A)

5/4

Answer 2

The limit can be interpreted as a Riemann sum for the integral $\int_0^1 f(x) dx$, where $f(x) = \lfloor \frac{2}{x} \rfloor - 2\lfloor \frac{1}{x} \rfloor$. The term $\lfloor 2y \rfloor - 2\lfloor y \rfloor$ is $1$ if $\{y\} \ge 0.5$ and $0$ otherwise. Thus, $f(x)=1$ when $\{\frac{1}{x}\} \ge 0.5$. This occurs for $x \in (\frac{1}{m+1}, \frac{1}{m+0.5}]$ for $m \ge 1$. The integral is $\sum_{m=1}^\infty (\frac{1}{m+0.5} - \frac{1}{m+1}) = 2\ln 2 - 1$. Comparing with $\log a - b$, we get $a=4, b=1$. Thus $\frac{a+b}{4} = \frac{4+1}{4} = \frac{5}{4}$.