# Sum of an Infinite GP

Consider the following sum:

\[S = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...\,\,\,{\rm{infinite\,\, terms}}\]

Will the sum be finite or infinite? This is a GP with the common ratio of magnitude less than 1. It turns out that all such GPs have finite sums. Let us find out the sum in this particular case.

We follow the same approach as earlier. We take the original series for *S*, and multiply it by the common ratio, and write the equal terms below each other. Then, we subtract the two series:

\[\begin{align}&S = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...\,\,\,{\rm{infinite \,\, terms}}\\&\frac{S}{2} = \,\,\,\,\,\,\,\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...\,\,\,{\rm{infinite\,\, terms}}\,\,\\ &\Rightarrow \,\,\,S - \frac{S}{2} = 1\,\,\, \Rightarrow \,\,\,\frac{S}{2} = 1\,\,\, \Rightarrow \,\,\,S = 2\end{align}\]

When we subtract the two series, only the first term is left, since by virtue of being infinite series, the other terms all cancel out.

Let us generalize this. Consider a GP with first term equal to *a* and common ratio *r* such that \(\left| r \right| < 1\). The GP will have a finite sum (this can be proved rigorously, but we won’t go into that here). Let us represent the sum (for infinitely many terms) of this series by \({S_\infty }\). We have:

\[\begin{align}&\;\;{S_\infty } = \,\,\,\,\,\,\,\,\,\,\, a + ar + a{r^2} + a{r^3}...\,\,\,{\rm{to }}\,\,\infty \\&r{S_\infty } = \,\,\,\,\,\,\,\,\,\,\,\,ar + a{r^2} + a{r^3}...\,\,\,{\rm{to }}\,\,\infty \\ &\Rightarrow \,\,\,{S_\infty } - r{S_\infty } = a\\ &\Rightarrow \,\,\,\left( {1 - r} \right){S_\infty } = a\,\,\, \Rightarrow \,\,\,{S_\infty } = \frac{a}{{1 - r}}\end{align}\]

**Example 1:** Find the following sums:

(a) \(S = \frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + ...\,\,{\rm{to }}\;\infty \)

(b) \(S = 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + ...\,\,\,{\rm{to }}\;\infty \)

**Solution:** (a) We have:

\[a = \frac{1}{3},\,\,\,r = \frac{1}{3}\]

Thus,

\[S = \frac{a}{{1 - r}} = \frac{{\frac{1}{3}}}{{1 - \frac{1}{3}}} = \frac{1}{2}\]

(b) In this case, we have:

\[a = 1,\,\,\,r = - \frac{1}{2}\]

Thus,

\[S = \frac{a}{{1 - r}} = \frac{1}{{1 - \left( { - \frac{1}{2}} \right)}} = \frac{2}{3}\]

**Example 2:** Find the following sum:

\[S = \frac{1}{3} + \frac{1}{{{5^2}}} + \frac{1}{{{3^3}}} + \frac{1}{{{5^4}}} + \frac{1}{{{3^5}}} + \frac{1}{{{5^6}}} + ...\,\,\,{\rm{to }}\infty \]

**Solution:** We can divide the given series into two separate GPs:

\[\begin{align}&{S_1} = \frac{1}{3} + \frac{1}{{{3^3}}} + \frac{1}{{{3^5}}} + ... = \frac{{\frac{1}{3}}}{{1 - \frac{1}{{{3^2}}}}} = \frac{{\frac{1}{3}}}{{\frac{8}{9}}} = \frac{3}{8}\\&{S_2} = \frac{1}{{{5^2}}} + \frac{1}{{{5^4}}} + \frac{1}{{{5^6}}} + ... = \frac{{\frac{1}{{{5^2}}}}}{{1 - \frac{1}{{{5^2}}}}} = \frac{{\frac{1}{{25}}}}{{\frac{{24}}{{25}}}} = \frac{1}{{24}}\end{align}\]

Thus,

\[S = {S_1} + {S_2} = \frac{3}{8} + \frac{1}{{24}} = \frac{5}{{12}}\]