Sum of a geometric series

Text

Compute the sum of the geometric series $\displaystyle\sum_{n=1}^{+\infty} 6\left(\frac{1}{3}\right)^{n}$.

Solution

The common ratio is $q=\tfrac13$, with $|q|<1$, so the series converges. The first term (for $n=1$) is $a_1=6\cdot\tfrac13=2$. The sum of a convergent geometric series is $S=\dfrac{a_1}{1-q}$: $S=\frac{2}{1-\tfrac13}=\frac{2}{\tfrac23}=3.$ $\boxed{\,S=3\,}$