This is a geometric progression
So
The general formUla is
[tex]\\ \rm\Rrightarrow a(n)=ar^{n-1}[/tex]
[tex]\\ \rm\Rrightarrow a(n)=\dfrac{1}{4}\left(\dfrac{1}{4}\right)^{n-1}[/tex]
Answer:
[tex]\displaystyle \lim_{n \to \infty} \dfrac{1}{3}\left(1-\left(\dfrac{1}{4}\right)^n\right)[/tex]
Step-by-step explanation:
Series: the sum of the elements of a sequence.
Therefore, as the numbers have been defined as a series and we need to find [tex]S_n[/tex]:
[tex]\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}+...[/tex]
First determine if the sequence is arithmetic or geometric.
If it is an arithmetic sequence, there will be a common difference between consecutive terms.
if it is a geometric sequence, there will be a common ratio between consecutive terms.
From inspection of the terms, we can see that there is a common ratio of 1/4, as each term is the previous term multiplied by 1/4, so it is a geometric series.
Sum of the first n terms of a geometric series:
[tex]S_n=\dfrac{a(1-r^n)}{1-r}[/tex]
Given:
[tex]a=\dfrac{1}{4}[/tex]
[tex]r=\dfrac{1}{4}[/tex]
Substitute the values of a and r into the formula:
[tex]\implies S_n=\dfrac{\frac{1}{4}\left(1-\left(\frac{1}{4}\right)^n\right)}{1-\frac{1}{4}}[/tex]
[tex]\implies S_n=\dfrac{\frac{1}{4}\left(1-\left(\frac{1}{4}\right)^n\right)}{\frac{3}{4}}[/tex]
[tex]\implies S_n=\dfrac{1}{3}\left(1-\left(\dfrac{1}{4}\right)^n\right)[/tex]
Therefore:
[tex]\displaystyle \lim_{n \to \infty} \dfrac{1}{3}\left(1-\left(\dfrac{1}{4}\right)^n\right)[/tex]
