Question

The value of $\underset{n\to \infty }{lim}\sum _{k=1}^n\log {\left(1+\frac{K}{n}\right)}^{\frac{1}{n}}$ is

• A. ${\log }_e\left(\frac{e}{4}\right)$
• B. ${\log }_e\left(\frac{4}{e}\right)$
• C. ${\log }_{e}4$
• D. None if these

Answer 1

The correct option is B ${\log }_e\left(\frac{4}{e}\right)$

${lim}_{n\to \infty }{\sum }_{k=1}^n\log {(1+\frac{k}{n})}^{\frac{1}{n}}$

$={lim}_{n\to \infty }{\sum }_{k=1}^n\frac{1}{n}\log (1+\frac{k}{n})[\therefore \log (x^m)=n\log x]$

$={lim}_{n\to \infty }\frac{1}{n}{\sum }_{k=1}^n\log (1+\frac{k}{n})$

$={\int }_0^1\log (1+x)dx[\therefore {lim}_{r=1}^{n}f(\frac{r}{n})={\int }_0^{1}f(x\left)dx\right]$

$=\left[\log \right(1+x)\int dx{]}_0^1-{\int }_0^1\left(\frac{d}{dx}log\right(1+x)\int dx).dx$

$=\left[x\log \right(1+x){]}_0^1-{\int }_0^1\frac{x}{1+x}.dx$

$=(1.\log 2-0)-{\int }_0^1\frac{(1+x)-1}{(1+x)}dx$

$=\log 2-{\int }_0^{1}dx+{\int }_0^1\frac{dx}{1+x}$

$=\log 2-(1-0)+\left[\log \right(1+1)-\log (1+0\left)\right]$

$=2\log 2-1=\log (2{)}^2-\log e=\log 4-\log e=\log \frac{4}{e}$

$\therefore {lim}_{n\to \infty }{\sum }_{k=1}^n\log {(1+\frac{k}{n})}^{\frac{1}{n}}=\log \frac{4}{e}$