Question

The value of $\underset{n\to \infty }{lim}\frac{1}{(n+1)}+\frac{1}{(n+2)}+\frac{1}{(n+3)}+...+...\frac{1}{(n+n)}=?$

• A. $log4$
• B. $log2$
• C. $log3$
• D. $log5$

Answer 1

The correct option is B $log2$

${lim}_{n\to \infty }(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+........+\frac{1}{n+n})={lim}_{n\to \infty {\sum }_{k=1}^n}\left(\frac{1}{n+k}\right)$
Choose the interval $[0,1]$ and divide in equal parts $[\frac{k-1}{n},\frac{k}{n}]$
${lim}_{n\to \infty }\left({\sum }_{k=1}^n\frac{1}{1+c_n}△t\right)$ where $c_n=\frac{k}{n}$ and $△t=\frac{1}{n}$ be the size of every subinterval
The limit of sum defines by the intergral
${lim}_{n\to \infty }\left({\sum }_{k=1}^{n}f\right(x\left)△x\right)={\int }_a^{b}f\left(x\right)dx$
So,
${\int }_0^1\frac{1}{1+t}=ln(1+1)-ln\left(1\right)=ln2$