Question

Calculate the sum of series $\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+.........$ upto infinity.

• A. $\frac{3}{4}$
• B. $\frac{5}{4}$
• C. $\frac{1}{4}$
• D. none of these

Answer 1

Answer: (A)

$\frac{3}{4}$

$S=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+....\infty \underset{–}{-\frac{S}{3}=\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+....\infty }\frac{2}{3}S=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+....\infty$

$\frac{2}{3}S=\frac{1/3}{(1-\frac{1}{3})}=\frac{1}{2}S=\frac{3}{4}$