# Series - $\sum_{i=1}^\infty (\frac{5}{12})^i$ - geometric series?

I have to solve - $$\sum_{i=1}^\infty \left(\frac{5}{12}\right)^i$$ - geometric series?

The geometric series sequence I know is - $$\sum_{i=0}^\infty x_i= \frac{1}{1-x}$$

However in my assignment, the series starts from $$i=1$$.

The solution I have is - $$\sum_{i=1}^\infty \left(\frac{5}{12}\right)^i = \frac{1}{1-\frac{5}{12}}-1$$

Can you explain please why is that the solution?

• Hint: Write out the first few terms. – Botond Jan 16 at 13:57

HINT: $$\sum_{i=0}^\infty x_i= \frac{1}{1-x} =x_0 + \sum_{i=1}^\infty x_i$$

• which means $\frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks. – Alan Jan 16 at 14:01
• exactly correct @Alan – Ahmad Bazzi Jan 16 at 14:01
There is another, faster, explanation, directly linked to the nature of the series: you can factor out $$x$$ from each term of the series: $$\sum_{i=1}^\infty \left(\frac{5}{12}\right)^{\mkern-5mu i}=x\sum_{i=1}^\infty \left(\frac{5}{12}\right)^{\mkern-5mu i-1}= x \sum_{i=0}^\infty \left(\frac{5}{12}\right)^{\mkern-5mu i}$$ (setting $$\;i\leftarrow i-1$$), so $$\sum_{i=1}^\infty \left(\frac{5}{12}\right)^{\mkern-5mu i}=\frac x{1-x}.$$
Observe that for suitable $$x$$: $$(1-x)\sum_{i=k}^{\infty}x^i=\sum_{i=k}^{\infty}x^i-\sum_{i=k+1}^{\infty}x^i=x^k$$so that:$$\sum_{i=k}^{\infty}x^i=\frac{x^k}{1-x}$$It remains to substitute $$k=1$$ and $$x=\frac5{12}$$.