Question

lf $S_n=[\frac{1}{1+\sqrt{n}}+\frac{1}{2+\sqrt{2n}}+\dots +\frac{1}{n+\sqrt{n^2}}]$ then $\underset{n\to \infty }{lim}{S}_n=$

• A. $\ln 2$
• B. $\ln 4$
• C. $\ln 8$
• D. $\ln 5$

Answer 1

The correct option is B $\ln 4$

The given series can be written as $\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{r+\sqrt{rn}}=\underset{n\to \infty }{lim}\frac{1}{n}\sum \frac{1}{r/n+\sqrt{r/n}}$
.
Converting the infinite limit into definite integral, we get
${\int }_0^1\frac{1}{x+\sqrt{x}}dx={\int }_0^1\frac{dx}{\sqrt{x}(\sqrt{x}+1)}$
.
Now substitute $\sqrt{x}+1=t\Rightarrow \frac{1}{2\sqrt{x}}dx=dt$
.
We get the integral as ${\int }_1^2\frac{2dt}{t}=2{\left[\ln t\right]}_1^2=2\ln 2=\ln 4$