How do I sum the series sum_(n=1)^oon(3/4)^n?

Feb 12, 2015

First of all: "How much is the sum of a geometric series?".

${\sum}_{n = 0}^{+ \infty} {r}^{n} = {a}_{0} / \left(1 - r\right)$, where r (such as $| r | < 1$) is called "ratio" ($r = {a}_{k} / {a}_{k - 1}$) and ${a}_{0}$ is the first term of the series (in this case ${a}_{0} = {r}^{0} = 1$).

Let's list all the terms of the series:

$1 \cdot \frac{3}{4} + 2 \cdot {\left(\frac{3}{4}\right)}^{2} + 3 \cdot {\left(\frac{3}{4}\right)}^{3} + 4 \cdot {\left(\frac{3}{4}\right)}^{4} + \ldots$, but we can also say:

$\frac{3}{4} + \frac{9}{16} + \frac{9}{16} + \frac{27}{64} + \frac{27}{64} + \frac{27}{64} + \frac{81}{256} + \frac{81}{256} + \frac{81}{256} + \frac{81}{256} + \ldots$,

or, better:

$\frac{3}{4} + \frac{9}{16} + \frac{27}{64} + \frac{81}{256} + \ldots +$

$\frac{9}{16} + \frac{27}{64} + \frac{81}{256} + \ldots +$

$\frac{27}{64} + \frac{81}{256} + \ldots +$

$\frac{81}{256} + \ldots +$

$\ldots$

The first sum is:

${\sum}_{n = 1}^{+ \infty} {\left(\frac{3}{4}\right)}^{n}$,

The second sum is:

${\sum}_{n = 2}^{+ \infty} {\left(\frac{3}{4}\right)}^{n}$,

The third sum is:

${\sum}_{n = 3}^{+ \infty} {\left(\frac{3}{4}\right)}^{n}$,

The fourth sum is:

${\sum}_{n = 4}^{+ \infty} {\left(\frac{3}{4}\right)}^{n}$,

$\ldots$

So:

${\sum}_{n = 1}^{+ \infty} n {\left(\frac{3}{4}\right)}^{n} = {\sum}_{n = 1}^{+ \infty} {\left(\frac{3}{4}\right)}^{n} + {\sum}_{n = 2}^{+ \infty} {\left(\frac{3}{4}\right)}^{n} + {\sum}_{n = 3}^{+ \infty} {\left(\frac{3}{4}\right)}^{n} + {\sum}_{n = 4}^{+ \infty} {\left(\frac{3}{4}\right)}^{n} + \ldots =$.

$= \frac{3}{4} \cdot \frac{1}{1 - \frac{3}{4}} + \frac{9}{16} \cdot \frac{1}{1 - \frac{3}{4}} + \frac{27}{64} \cdot \frac{1}{1 - \frac{3}{4}} + \frac{81}{256} \cdot \frac{1}{1 - \frac{3}{4}} + \ldots =$

$= \frac{3}{4} \cdot 4 + \frac{9}{16} \cdot 4 + \frac{27}{64} \cdot 4 + \frac{81}{256} \cdot 4 + \ldots =$

$= 4 \left(\frac{3}{4} + \frac{9}{16} + \frac{27}{64} + \frac{81}{256} + \ldots\right) =$

$= 4 \cdot {\sum}_{n = 1}^{+ \infty} {\left(\frac{3}{4}\right)}^{n} = 4 \cdot \frac{3}{4} \cdot \frac{1}{1 - \frac{3}{4}} = 4 \cdot \frac{3}{4} \cdot 4 = 12$.