Simplifying the infinite series $\sum_{n = 1}^{\infty} \left(\frac{1}{2}\right)^{3n}$

Question

Does anyone know of a way to simplify

$$ \sum_{n = 1}^{\infty} \left(\frac{1}{2}\right)^{3n} $$

to a number?

Answer 1

Notice that we all know $ \sum_{k=1} a^k = \frac{a}{1-a} $ if $a \in (-1,1) $.

Applying this trick to your problem:

$$ \sum \frac{1}{2^{3n}} = \sum \left(\frac{1}{2^3}\right)^n = \frac{1/2^3}{1 - 1/2^3}= {\frac{1}{8}\over1-\frac{1}{8}}=\frac{1}{7}$$

Answer 2

its summation to infinity of a geometric series whose first term (a) is 1/8 and the common ratio (r) is also 1/8. Sum to infinity is then a/(1-r). Substituting, we get (1/8)/(1-1/8) which simplifies to 1/7.