find-the-sum-n-1-1-2-n-i-1-3-n-

Question Number 165262 by mkam last updated on 28/Jan/22

f i n d t h e s u m \underset{n = 1}{\sum^{\infty}} {(\frac{1}{2})}^{n} + i {(\frac{1}{3})}^{n}

Answered by Mathspace last updated on 28/Jan/22

\sum_{n = 1}^{\infty} {(\frac{1}{2})}^{n} + i {(\frac{1}{3})}^{n}} = \sum_{n = 0}^{\infty} {(\frac{1}{2})}^{n} - 1

+ i \sum_{n = 0}^{\infty} {(\frac{1}{3})}^{n} - i

= \frac{1}{1 - \frac{1}{2}} - 1 + i \frac{1}{1 - \frac{1}{3}} - i

= 1 + i \frac{3}{2} - i = 1 + i (\frac{3}{2} - 1)

= 1 + \frac{1}{2} i

Answered by Ar Brandon last updated on 28/Jan/22

S = \underset{n = 1}{\sum^{\infty}} [{(\frac{1}{2})}^{n} + i {(\frac{1}{3})}^{n}]

= \underset{n = 1}{\sum^{\infty}} {(\frac{1}{2})}^{n} + i \underset{n = 1}{\sum^{\infty}} {(\frac{1}{3})}^{n}

= 1 + i \frac{1}{2}