Question

Sum $1+\frac{3}{2}+\frac{5}{4}+\frac{7}{8}+.....$ to infinity.

Answer 1

Let $S=1+\frac{3}{2}+\frac{5}{4}+\frac{7}{8}......\infty$ ....... $\left(1\right)$
Multiplying both sides by $\frac{1}{2}$, we get
$S\frac{1}{2}=0+\frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}......\infty$ ......... $\left(2\right)$
Subtracting $\left(1\right)$ by $\left(2\right),$ we get
$S\frac{1}{2}=1+\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\frac{2}{16}.............\infty$
After unit above series is infinite G.P. with common ratio $=\frac{1}{2}$
Then, $S\frac{1}{2}=1+2\times \frac{\frac{1}{2}}{1-\frac{1}{2}}=3$
$\therefore S=6$