An Algebraic Limit

This sum equals:

$S = 3( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots)$ , which is the geometric sum

$S = \sum_{k=1}^{\infty} 3*2^{-k}$ .

Recall that the sum of a geometric series $S = (a + ar + ar^2 + \dots)$ is given by the formula $S = \frac{a}{1-r}$ .

In this case $a = \frac{3}{2}, r = \frac{1}{2}$ and $1-r = \frac{1}{2}$ , thus $S = \frac{\frac{3}{2}}{\frac{1}{2}} = 3$ .

For more about geometric series and sums, including the derivation of the formula for the sum, see the Brilliant page on the topic.

$\begin{aligned} S & = \left(\frac 12+\frac 12+ \frac 12 \right) + \left(\frac 14+\frac 14+ \frac 14 \right) + \left(\frac 18+\frac 18+ \frac 18 \right) + \cdots \\ & = 3 \left(\frac 12 + \frac 14 + \frac 18 + \cdots \right) \\ 2S & = 6 \left(\frac 12 + \frac 14 + \frac 18 + \cdots \right) \\ & = 3 \left(1+\frac 12 + \frac 14 + \frac 18 + \cdots \right) \\ & = 3 + 3 \left(\frac 12 + \frac 14 + \frac 18 + \cdots \right) \\ \implies 2S & = 3 + S \\ S & = \boxed 3 \end{aligned}$

Note We have assumed that the sum $\frac 12 + \frac 14 + \frac 18 + \cdots$ converges in the above calculations. The sum does converge to $\frac 12 \left(\frac 1{1-\frac 12} \right) = 1$ , which also leads to $S = \boxed 3$ .

Let a = the sum of the first bracket the sum of the second bracket is a(1/2), the sum of the third bracket is a(1/4). a(1/2) + a(1/4) + a(1/8)... approaches a(1). Here we have solved for every bracket from the second one and onwards. Adding the first bracket, the answer is 2a. a = 1.5 therefore 2a = 3, which is the answer