Question

Let $a,b,cϵR^{+}$ and $\underset{n\to \infty }{lim}\sum _{k=1}^n\frac{n}{(k+an)(k+bn)}=\frac{A}{a-b}ln\frac{a(b+B)}{b(a+C)},a\ne b,$ then $(A+B+C)$ is equal to

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

The correct option is B 3

Let $a,b,c$ $ϵR^{+}$and
${lim}_{n\to \infty }{\sum }_{k=1}^n\frac{n}{n^2(a+\frac{k}{n})(b+\frac{k}{n})}$
$={lim}_{n\to \infty }\frac{1}{n}{\sum }_{k=1}^n\frac{1}{(a+\frac{k}{n})(b+\frac{k}{n})}$
$={\int }_0^1\frac{1}{(a+x)(b+x)}dx=\frac{1}{a-b}{\int }_0^1\frac{(a+x)-(b+x)}{(a+x)(b+x)}dx$
$=\frac{1}{a-b}{\int }_0^1(\frac{1}{b+x}-\frac{1}{a+x})dx$
$=\frac{1}{a-b}{\left(ln\right(b+x)-ln(a+x\left)\right)}_0^1$
$=\frac{1}{a-b}{\left(ln\frac{(b+x)}{(a+x)}\right)}_0^1=\frac{1}{a-b}ln\frac{(b+1)a}{(a+1)b}$
$\therefore A+B+C=3$
Hence, option 'B' is correct.