Verified Answer :

mackiemesser

\(1/2+1/4+1/8+1/16+...=\sum\limits_{n=1}^\infty (\frac{1}{2})^n = \frac{\frac{1}{2}}{1-\frac{1}{2}}=1\)

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:

\(S_\infty = \frac{a}{1-r}\) ......[1] ;where \(0

Given the series: \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.....\)

Since, this series is geometric series with constant term(r) = \(\frac{1}{2}\)

Since,

\(\frac{\frac{1}{4}}{\frac{1}{2} } =\frac{2}{4} = \frac{1}{2}\),

\(\frac{\frac{1}{8}}{\frac{1}{4}} =\frac{4}{8} = \frac{1}{2}\) and so on....

Here, first term(a) = \(\frac{1}{2}\)

Substitute the values of a and r in [1] we get;

\(S_\infty = \frac{\frac{1}{2}}{1-\frac{1}{2}}\)   where r =  \(\frac{1}{2}< 1\)

\(S_\infty = \frac{\frac{1}{2}}{\frac{2-1}{2}}\)  

or

\(S_\infty = \frac{\frac{1}{2}}{\frac{1}{2}}\)

Simplify:

 \(S_\infty = 1\)  

Therefore, the sum of the infinite geometric series is, 1

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