Geometric Series & Repeating Decimals

[KWF]

How is a/(1-r) determined from the following?

When r < 1 then, a + ar + ar^2 + ar^3 + .... = a / (1-r)

Evaluating the repeating decimal 0.27272727... using geometric
series, we have

0.272727... = 0.27 + 0.0027 + 0.000027 + 0.00000027 + ...
= 0.27 + 0.27(.01) + 0.27(.01)^2 + 0.27(.01)^3 + ...
= 0.27 / (1-.01)
= 0.27 / 0.99
= 27/99
= 3/11
Likewise, how is 0.27 = (1-.01) determined from the calculation above? See the third line. I thank you for your reply!

 

[Bob Brown MSEE]

How is a/(1-r) determined from the following?

When r < 1 then, a + ar + ar^2 + ar^3 + .... = a / (1-r)

Evaluating the repeating decimal 0.27272727... using geometric
series, we have

0.272727... = 0.27 + 0.0027 + 0.000027 + 0.00000027 + ...
= 0.27 + 0.27(.01) + 0.27(.01)^2 + 0.27(.01)^3 + ...
= 0.27 / (1-.01) <==== a / (1-r)
= 0.27 / 0.99
= 0.27/99
= 0.03/11
Likewise, how is 0.27 = (1-.01) determined from the calculation above? See the third line. I thank you for your reply!

r=0.01
a=0.27

so: a / (1-r) = 0.27 / (1-.01)

 

[stapel]

When r < 1 then, a + ar + ar^2 + ar^3 + .... = a / (1-r)

WHO told you that?

a + ar + a^2 + ar^3 + ..... +ar^(n) = a[r^(n+1) - 1] / (r - 1)

Note difference in formulas (in particular, the difference in the "number" of terms):

. . . . .\(\displaystyle \mbox{1) }\displaystyle{\, a\, +\, ar\, +\, a^2\, +\, ar^3\, +\, ...\, =\, \sum_{i\, =\, 0}^{\infty}\, ar^i\, =\, \dfrac{a}{1\, -\, r}}\)

. . . . .\(\displaystyle \mbox{2) }\displaystyle{\, a\, +\, ar\, +\, a^2\, +\, ar^3\, +\, ...\, +\, ar^n\, =\, \sum_{i\, =\, 0}^{n}\, ar^i\, =\, \dfrac{a\left(1\, -\, r^{n+1}\right)}{1\, -\, r}}\)

The form of the last term -- indeed, the existence of a "last" term -- greatly changes the result!

 

[eddybob123]

How is a/(1-r) determined from the following?

When r < 1 then, a + ar + ar^2 + ar^3 + .... = a / (1-r)

This is the sum of an infinite geometric sequence:

\(\displaystyle a+ar+ar^2+ar^3+...=S\)

\(\displaystyle ar+ar^2+ar^3+ar^4=Sr\)

\(\displaystyle a=S-Sr\)

\(\displaystyle a=S(1-r)\)

\(\displaystyle S=\frac{a}{1-r}\)