Using Mathematica, I have been able to make the following statement based on numerical evidence: $$\sum_{i=0}^\infty \frac{2^i}{x^i}=\frac{x}{x-2}$$ for any $x≥3$. How can this be proven?
Asked
Active
Viewed 96 times
1
-
This is a geometric series. – user84413 Feb 21 '16 at 05:19
-
See http://math.stackexchange.com/questions/370662/infinite-geometric-series-formula-derivation – GGG Feb 21 '16 at 05:21
2 Answers
3
Define $r = 2/x$. Then $$\begin{align*} \sum_{i=0}^\infty \frac{2^i}{x^i} = \sum_{i=0}^\infty r^i = \frac{1}{1-r} = \frac{x}{x-2} \end{align*}$$ where the summation formula for geometric series is used.
Henricus V.
- 18,694
1
This can solved from scratch by taking one term out of the summation, changing the index range and refactoring to obtain $$ \begin{align} S= \sum_{i=0}^{\infty} \left(\frac{2}{x}\right)^i &= 1+\sum_{i=1}^{\infty} \left(\frac{2}{x}\right)^i \\ &= 1+\sum_{i=0}^{\infty} \left(\frac{2}{x}\right)^{i+1} \\ &= 1+\frac{2}{x}\sum_{i=0}^{\infty} \left(\frac{2}{x}\right)^{i} \\ \end{align}$$
Now grouping the series terms yields $$ \begin{align} \left(1-\frac{2}{x}\right)\sum_{i=0}^{\infty} \left(\frac{2}{x}\right)^i &= 1 \\ \sum_{i=0}^{\infty} \left(\frac{2}{x}\right)^i &=\frac{1}{1-\frac{2}{x}}=\frac{x}{x-2} \\ \end{align}$$
where the sum converges for $\vert \frac{2}{x} \vert\lt 1$, not only $x \ge 3$.
Are you assuming positive integer $x$?
Jaume Oliver Lafont
- 5,212