Question

Evaluate: $\underset{n\to \infty }{lim}\sum _{r=0}^{n-1}\frac{1}{n+r}$

• A. $\log 2$
• B. $2\log 2$
• C. $\frac{1}{2}\log 2$
• D. $\frac{1}{4}\log 2$

Answer 1

The correct option is A $\log 2$

$\underset{n\to \infty }{lim}\sum _{r=0}^{n-1}\frac{1}{n+r}$

$=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=0}^{n-1}\frac{1}{(1+\frac{r}{n})}$

Now, when $r=0,\frac{r}{n}=0$

and when $r=n-1,\underset{n\to \infty }{lim}\frac{r}{n}=1$

Now, the given summation can be written as-

${\int }_0^1\frac{1}{1+x}dx$

$={\left[\log (1+x)\right]}_0^1$

$=\log 2-\log 1=\log 2$

Hence, the answer is option-(A).