sponges12
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What is the sum of the infinite geometric series?.

1/2+1/4+1/8+1/16+...

Sagot :

mackiemesser
[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 :)

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

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