Finding the infinite sum of a geometric series

Question

I have the series: $3 + \frac{3}{2} + \frac{3}{4} + \frac{3}{8}$,

I need to calculate the exact value of it. I have found that the equation of this is $a_n = 3(\frac{1}{2})^{n-1}$, but I am not sure how to progress from here. Would guess that I need to convert it to sigma notation but I am not totally sure.

$$\sum_{n=1}^{\infty}ax^{n-1} = \dfrac{a}{1-x}$$ – David P Dec 21 '18 at 04:25 — David P, Dec 21 '18 at 04:25

score 2 · Accepted Answer · answered Dec 21 '18 at 04:25

2

You have the terms right: you're sum is $$\sum_{n=1}^\infty 3\left(\frac{1}{2}\right)^{n-1}=3\sum_{n=0}^\infty\left(\frac{1}{2}\right)^n$$ Now recall the geometric series formula is $$\sum_{n=0}^\infty x^n=\frac{1}{1-x}\qquad |x|<1$$ So what is $x$ in your case?

answered Dec 21 '18 at 04:25

Dave

13,568

$x = \frac{1}{2}$ which would make the answer $2$? – Big Papa Dec 21 '18 at 04:27
Yes, but don't forget you're multiplying the geometric series by $3$. – Dave Dec 21 '18 at 04:28
1

Ohhh, making the final answer 6? Thanks for the help! – Big Papa Dec 21 '18 at 04:30

score 0 · Answer 2 · answered Dec 21 '18 at 04:27

I'm going to assume this is supposed to go on forever. First, I'm going to solve this problem, then show the general case.

Suppose $S = 3 + \frac{3}{2} + \frac{3}{4} + \cdots$. Note that $\frac{S}{2} = \frac{3}{2} + \frac{3}{4} + \frac{3}{8} + \cdots$ Now, we note that $S - \frac{S}{2} = 3$, because the second sequence is missing nothing but the first term. We can then solve to show that $S = \boxed{6}$.

Now, note we have the general geometric sequence $S = a + a \times x + a \times x^2 + \cdots$. We know that $S \times x = a \times x + a \times x^2 + a \times x^3 + \cdots \rightarrow S - S \times x = a \rightarrow S = \boxed{\frac{a}{1-x}}$. Note this only works for $|x| < 1$, or else the geometric sequence would not converge.

Finding the infinite sum of a geometric series

2 Answers2