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.
On
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.
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?