Question

For all $n\in N,1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots +\frac{1}{\sqrt{n}}=S_n$,then $S_n$ is

• A. $\sqrt{n}$
• B. $<\sqrt{n}$
• C. $\le \sqrt{n}$
• D. $\ge \sqrt{n}$

Answer 1

The correct option is C $\le \sqrt{n}$

For any $r\le n$ , $\frac{1}{\sqrt{r}}\ge \frac{1}{\sqrt{n}}$
We approximate each term in the series by a smaller term $\frac{1}{\sqrt{n}}$
Hence,
$S_n\le n\times \frac{1}{\sqrt{n}}$