Question

Let $S_n=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+.....$ to $n$ terms, then $\underset{n\to \infty }{lim}{S}_n$ is equal to

• A. $\frac{1}{3}$
• B. $3$
• C. $\frac{1}{4}$
• D. $\infty$

Answer 1

The correct option is A $\frac{1}{3}$

$S_n=\frac{1}{1\times 4}+\frac{1}{4\times 7}+\frac{1}{7\times 10}+...\dots \dots \dots \dots \dots .$

$\underset{n\to \infty }{lim}\frac{1}{3}(\frac{3}{1\times 4}+\frac{3}{4\times 7}+\frac{3}{7\times 10}+...\dots \dots \dots ..)$

$\frac{3}{n\times (n+3)}=\frac{1}{n}-\frac{1}{n+3}$

$\underset{n\to \infty }{lim}\frac{1}{3}(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...\dots \dots \dots -\frac{1}{n+3})$

$\underset{n\to \infty }{lim}\frac{1}{3}(1-\frac{1}{n+3})=\frac{1}{3}$