Question

If $a,b,c\in R^{+},$ then $\underset{n\to \infty }{lim}\sum _{k=1}^n\frac{n}{(k+an)(k+bn)}$ is equal to

• A. $\frac{1}{a-b}\ln \left(\frac{b(b+1)}{a(a+1)}\right)$ if $a\ne b$
• B. $\frac{1}{a-b}\ln \left(\frac{a(b+1)}{b(a+1)}\right)$ if $a\ne b$
• C. non-existent if $a=b$
• D. $\frac{1}{a(1+a)}$ if $a=b$

Answer 1

Answer: (B), (D)

$\frac{1}{a(1+a)}$ if $a=b$

$\underset{n\to \infty }{lim}\sum _{k=1}^n\frac{n}{(k+an)(k+bn)}$
$=\underset{n\to \infty }{lim}\sum _{k=1}^n\frac{n}{n^2(a+\frac{k}{n})(b+\frac{k}{n})}$
$=\underset{n\to \infty }{lim}\sum _{k=1}^n\frac{1}{(a+\frac{k}{n})(b+\frac{k}{n})}\cdot \frac{1}{n}$
$=\underset{0}{\overset{1}{\int }}\frac{1}{(a+x)(b+x)}dx$
$=\frac{1}{a-b}\underset{0}{\overset{1}{\int }}\frac{(a+x)-(b+x)}{(a+x)(b+x)}dx$
$=\frac{1}{a-b}\underset{0}{\overset{1}{\int }}(\frac{1}{b+x}-\frac{1}{a+x})dx$
$=\frac{1}{a-b}{\left[\ln \left(\frac{b+x}{a+x}\right)\right]}_0^1$
$=\frac{1}{a-b}\ln \left(\frac{a(b+1)}{b(a+1)}\right)$


Now, if $a=b,$ then we have
$\underset{n\to \infty }{lim}\sum _{k=1}^n\frac{1}{{(a+\frac{k}{n})}^2}\cdot \frac{1}{n}$
$=\underset{0}{\overset{1}{\int }}\frac{1}{{(a+x)}^2}dx$
$={\left[\frac{-1}{a+x}\right]}_0^1=\frac{1}{a(a+1)}$