Question

Evaluate: $\sum _{r=1}^{\infty }\frac{r}{7^r}$

• A. $\frac{5}{36}$
• B. $\frac{7}{36}$
• C. $\frac{9}{35}$
• D. None of these

Answer 1

The correct option is B $\frac{7}{36}$

$s={\sum }_{r=1}^{\infty }\frac{r}{7^r}=\frac{1}{7}+\frac{2}{7^2}+\frac{3}{7^3}...$......(1)

$\frac{1}{7}s=\frac{1}{7^2}+\frac{2}{7^3}+\frac{3}{7^4}......$.....(2)

Eq,(1)-eq.(2)

$s-\frac{1}{7}s=\frac{1}{7}+\frac{1}{7^2}+..........$

Which is G.P. with $a=\frac{1}{7}=r$

$\therefore \frac{6s}{7}=\frac{\frac{1}{7}}{1-\frac{1}{7}}$

$s=\frac{7}{36}$