Question

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

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

Answer 1

Answer: (B)

$\frac{1}{a-b}\ln \frac{a\left(b+1\right)}{b\left(a+1\right)}ifa\ne b$

$a,b,c\in R^{+}$

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

$={lim}_{n\to \infty }{\sum }_n^{k=1}\frac{n}{n^2(\frac{k}{n}+a)(\frac{k}{n}+b)}$

Let $x=\frac{k}{n},dx=\frac{1}{n}$

Now limit of $x$ varies from $0$ to $1$

$={\int }_0^1\frac{1}{(x+a)(x+b)}dx$

$={\int }_0^1(\frac{1}{x+b}-\frac{1}{x+a})\frac{1}{a-b}dx$

$=\frac{1}{a-b}{\int }_0^1(\frac{1}{x+b}-\frac{1}{x+a})dx$

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

$=\frac{1}{a-b}\left[\log \right(1+b)-\log (1+a)-\log (1+a)-\log (b)+\log a]$

$=\frac{1}{a-b}\left[\log \left(\frac{a(1+b)}{b(1+a)}\right)\right]$; if $a\ne b$.

and non existent if $a=b$.