Answer :
[tex]1/2+1/4+1/8+1/16+...=\sum\limits_{n=1}^\infty (\frac{1}{2})^n = \frac{\frac{1}{2}}{1-\frac{1}{2}}=1[/tex]
We can see that consecutive fractions are made from 1/2 to consecutive powers. Because we begin with 1/2, n=1
We will infinitely add fractions , hence Lemniscate sign.
In other words you cannot find a number which needs to be added to the geometric series to get "1", therefore the answer is 1. I remember the teacher explaining it this way :)
We can see that consecutive fractions are made from 1/2 to consecutive powers. Because we begin with 1/2, n=1
We will infinitely add fractions , hence Lemniscate sign.
In other words you cannot find a number which needs to be added to the geometric series to get "1", therefore the answer is 1. I remember the teacher explaining it this way :)
Answer:
The sum of the given geometric series is, 1
Step-by-step explanation:
Geometric sequence states that a sequence of numbers that follows a pattern were the next term is found by multiplying by a constant called the common ratio (r).
The sum of the infinite terms of a geometric series is given by:
[tex]S_\infty = \frac{a}{1-r}[/tex] ......[1] ;where [tex]0<r<1[/tex]
Given the series: [tex]\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.....[/tex]
Since, this series is geometric series with constant term(r) = [tex]\frac{1}{2}[/tex]
Since,
[tex]\frac{\frac{1}{4}}{\frac{1}{2} } =\frac{2}{4} = \frac{1}{2}[/tex],
[tex]\frac{\frac{1}{8}}{\frac{1}{4}} =\frac{4}{8} = \frac{1}{2}[/tex] and so on....
Here, first term(a) = [tex]\frac{1}{2}[/tex]
Substitute the values of a and r in [1] we get;
[tex]S_\infty = \frac{\frac{1}{2}}{1-\frac{1}{2}}[/tex] where r = [tex]\frac{1}{2}< 1[/tex]
[tex]S_\infty = \frac{\frac{1}{2}}{\frac{2-1}{2}}[/tex]
or
[tex]S_\infty = \frac{\frac{1}{2}}{\frac{1}{2}}[/tex]
Simplify:
[tex]S_\infty = 1[/tex]
Therefore, the sum of the infinite geometric series is, 1
