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

Answer: (A)

$\frac{1}{3}$

$s_n=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...\Rightarrow s_n={\sum }_{n=1}^{\infty }\frac{1}{(3n-2)(3n+1)}={\sum }_{n=1}^{\infty }\frac{1}{3}(\frac{1}{3n-2}-\frac{1}{3n+1})$
Hence
$\underset{n\to \infty }{lim}{s}_n=\underset{n\to \infty }{lim}\sum _{n=1}^{\infty }\frac{1}{3}(\frac{1}{3n-2}-\frac{1}{3n+1})=\underset{n\to \infty }{lim}\left(\frac{1}{3}(1-\frac{1}{3n+1})\right)=\frac{1}{3}$
Hence, option 'A' is correct.