Answer:
[tex]\dfrac{1}{3}[/tex]
Step-by-step explanation:
Given:
[tex]\displaystyle \sum^{\infty}_{n=1} 4^{-n}[/tex]
The sigma notation means to find the sum of the given geometric series where the first term is when n = 1 and the last term is when n = ∞.
To find the first term in the series, substitute n = 1 into the expression:
[tex]\implies a_1=4^{-1}=\dfrac{1}{4}[/tex]
The common ratio of the geometric series can be found by dividing one term by the previous term:
[tex]\implies r=\dfrac{a_2}{a_1}=\dfrac{4^{-2}}{4^{-1}}=\dfrac{1}{4}[/tex]
As | r | < 1 the series is convergent.
When a series is convergent, we can find its sum to infinity (the limit of the series).
Sum to infinity formula:
[tex]S_{\infty}=\dfrac{a}{1-r}[/tex]
where:
Substitute the found values of a and r into the formula:
[tex]S_{\infty}=\dfrac{\frac{1}{4}}{1-\frac{1}{4}}=\dfrac{1}{3}[/tex]
Therefore:
[tex]\displaystyle \sum^{\infty}_{n=1} 4^{-n}=\dfrac{1}{3}[/tex]
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