1. Why cant you use the formula of the infinite geometric series on 1/81 + 1/27 + 1/9 + 1/3 + .....
2. Using concepts of infinite geometric series, show that 0.9999 ..... = 1.
r = (1/27) ÷ (1/81)
= 3
so the series diverges, that is, the terms are getting larger.
Think about adding terms that keep getting larger.
Clearly the sum will get larger and larger and eventually become infinitely large.
That is why for the formula
Sum (infinite number of terms) = a/(1-r)
part of the definition will be |r| < 1
2.
let .99999.. = .9 + .09 + .009 + ....
so a = .9 , and r = .1
sum(all terms) = .9/(1-.1)
= .9/.9
= 1
1. To use the formula for the sum of an infinite geometric series, the common ratio (r) must be between -1 and 1.
In the given sequence, the common ratio is 1/3. Since 1/3 is greater than 1, it does not meet the requirement of the formula. Therefore, we cannot directly use the formula to calculate the sum of the series 1/81 + 1/27 + 1/9 + 1/3 + ...
However, there is an alternative method to find the sum of this divergent series using a different approach, such as using partial sums or limit calculations.
2. To prove that 0.9999... (repeating) is equal to 1 using concepts of infinite geometric series, we consider the fraction form of the repeating decimal: 0.9999... = 9/10 + 9/100 + 9/1000 + ...
Let's denote the sum of this series as S.
Multiply both sides of the equation by 1/10 to get:
(1/10)*(0.9999...) = (1/10)*(9/10 + 9/100 + 9/1000 + ...)
Simplifying the right side, we have:
(1/10)*(0.9999...) = (9/100) + (9/1000) + (9/10000) + ...
Now, subtracting the original equation from the shifted equation:
(1/10)*(0.9999...) - 0.999... = [(9/100) + (9/1000) + (9/10000) + ...] - [0.9999...]
On the left side, we can simplify the subtraction:
(1/10)*(0.9999...) - 0.999... = (0.9999...) - 0.999...
The right side becomes an infinite geometric series, so we can apply the formula for the sum of an infinite geometric series with a common ratio of 1/100.
Using the formula for the sum of an infinite geometric series:
(1/10)*(0.9999...) - 0.999... = (0.9999...) - 0.999...
(1/10)*(0.9999...) - 0.999... = (9/100)/(1 - 1/100)
Simplifying the right side of the equation:
(1/10)*(0.9999...) - 0.999... = (9/100)/(99/100)
(1/10)*(0.9999...) - 0.999... = 9/99
Simplifying the left side of the equation:
0.1 * (0.9999...) - 0.999... = 1/11
Using the properties of decimals and fractions, we can simplify the equation further:
0.1 * (0.9999...) - 0.999... = 1/11
0.09999... - 0.999... = 1/11
0.9 = 1/11
Since 0.9 is equal to 1/11, it implies that:
1/11 = 0.9999...
Therefore, we can conclude that 0.9999... (repeating) is equal to 1.
1. The formula for the sum of an infinite geometric series is given by S = a / (1 - r), where "a" is the first term and "r" is the common ratio. However, in order to use this formula, the absolute value of the common ratio (|r|) must be less than 1.
In the given series 1/81 + 1/27 + 1/9 + 1/3 + ..., the common ratio is 1/3, which is greater than 1. Therefore, the absolute value of the common ratio is not less than 1, and we cannot use the formula for the sum of an infinite geometric series to find the sum of this sequence.
However, we can still find the sum of this series using a different approach, such as finding the limit of the partial sums.
2. To show that 0.9999... is equal to 1, we can use the concept of an infinite geometric series.
Let x = 0.9999...
Multiplying both sides of the equation by 10 gives:
10x = 9.9999...
Now, let's subtract the original equation from this new equation:
10x - x = 9.9999... - 0.9999...
Simplifying both sides of the equation gives:
9x = 9
Dividing both sides of the equation by 9 gives:
x = 1
Therefore, we have shown that 0.9999... is equal to 1 using the concept of an infinite geometric series.