The two major types of series are the arithmetic series and the geometric series. The arithmetic series is characterized by common difference, while the geometric series has common ratio between two successive terms.
The sum of the given series is:
[tex]\lim_{n \to \infty} \frac{1}{3}(1 - \frac{1}{4}^n )[/tex]
The given series is a geometric series. So, we first calculate the common ratio (r) using:
[tex]r = T_2 \div T_1[/tex]
From the series, we have:
[tex]T_1 = \frac{1}{4}[/tex]
[tex]T_2 = \frac{1}{16}[/tex]
So, the equation becomes
[tex]r = \frac{1}{16} \div \frac{1}{4}[/tex]
Rewrite as product
[tex]r = \frac{1}{16} * \frac{4}{1}[/tex]
[tex]r = \frac{1}{4}[/tex]
The formula to calculate the sum of a geometric series of is:
[tex]S_n = \frac{a(1 - r^n )}{1-r}[/tex]
Where
[tex]a = T_1 =\frac{1}{4}[/tex] -- the first term
[tex]S_n = \frac{\frac{1}{4}(1 - \frac{1}{4}^n )}{1-\frac{1}{4}}[/tex]
Simplify the denominator
[tex]S_n = \frac{\frac{1}{4}(1 - \frac{1}{4}^n )}{\frac{3}{4}}[/tex]
Divide 1/4 by 3/4
[tex]S_n = \frac{1}{3}(1 - \frac{1}{4}^n )[/tex]
We can conclude that, the sum of the series is:
[tex]S_n = \lim_{n \to \infty} \frac{1}{3}(1 - \frac{1}{4}^n )[/tex]
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