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

Question

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?

Answer 1

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

Answer 2

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 $$\sum_{i=1}^\infty \left(\frac{5}{12}\right)^{\mkern-5mu i}=\frac x{1-x}.$$

Answer 3

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 $x=\frac5{12}$.