Calculate sum of $\sum \limits_{n=1}^{\infty} \frac{3}{7^n} $

Question

I need help with this excersise, can't figure it out, thanks alot !

Answer 1

Hint: $\sum_{n\in\mathbb N}\frac{3}{7^n}=3\sum_{n\in\mathbb N}\left(\frac{1}{7}\right)^n$, which is a geometric series.

Answer 2

use the geometric series for $|x|<1$ $$1+x+x^2+x^3+.....=\frac{1}{1-x}$$ for your series $$x+x^2+x^3+x^4+....=x(1+x+x^2+x^3+....)=x*\frac{1}{1-x}$$ use $x=1/7$ $$3*(1/7)*\frac{1}{1-1/7}=\frac{1}{2}$$

Answer 3

This is a simple infinite geometric series with common ratio (r) 1/7 and first term (a) 3/7. The sum = a/(1-r)

Answer 4

$$\sum\limits_{n=1}^{\infty} \frac{3}{7^n} =3\sum\limits_{n=1}^{\infty} \frac{1}{7^n} =3\sum\limits_{n=1}^{\infty} \left|\frac{1}{7}\right|^n$$ Since $\left|\frac17\right| \lt 1$, then $$3\sum\limits_{n=1}^{\infty} \left|\frac{1}{7}\right|^n=3\left(\frac{\frac17}{1-\frac17}\right)$$ $$=3\left(\frac{1}{7-1}\right)=\frac{3}{6}=\frac12$$ Study geometric series.