Math 11

Week 2 Geometric Series

February 13, 2018 jessicap2015 Leave a comment

Week 2: This week in Pre Calculus 11 we learned about infinite and finite geometric series.

GEOMETRIC SEQUENCES – Each term is multiplied by a constant, known as the common ratio.

There are two types of infinite geometric series. (adding the sequence is a series to find a sum).

DIVERGING – A diverging series means the ratio is r > 1 or r < -1. This means the number of terms will increase, which means the partial sum increases, so the series does not have a finite sum. (NO SUM) It only has a sum if it is asking for a specific sum of terms.

CONVERGING – A converging series means the ratio is 1 > r > 0 (smaller than 1, bigger than 0) or -1 < r < 0. ( bigger that -1, smaller than 0). The partial sum will appear to get closer to a number, therefore we estimate the finite sum. (HAS SUM).

To find the sum:

Find Ratio
Identify if it Diverges or Converges.
Diverges, STOP (No Sum)
Converges, CONTINUE (Sum)
Use the formula $S_{\infty} = \frac {a}{1- r}$

ex. 8 + 16 + 32 + 64 + 128 + 256 …

Ratio: $\frac {16}{8} = 2$

Identify: Diverges. NO SUM

ex. 2) 8 + -7.2 + 6.48 + -5.832 + 5.2488 …

Ratio: $\frac {-7.2}{8} = -0.9$

Identify: Converges, ratio is a decimal bigger that -1 but smaller than 0.

FORMULA :

$S_{\infty} = \frac {a}{1 - r}$
$S_{\infty} = \frac {8}{1 - (-0.9)}$
$S_{\infty} = \frac {8}{1 + 0.9}$
$S_{\infty} = \frac {8}{1.9}$
$S_{\infty}$ = 4.21052…
Estimated sum = 4.2

This is how to find the sum of a geometric series.

If a question asks for the sum of a term the formula to use is:

$S_{n} = a \frac {(r^n -1)}{r - 1}$

ex. 8 + 16 + 32 + 64 + 128 + 256 …

Find $S_{10}$ :

$S_{10} = a \frac {(r^n -1)}{r - 1}$
$S_{10} = 8 \frac {(2^{10} - 1)}{2 - 1}$
$S_{10} = 8 \frac {(1024-1)}{2-1}$
$S_{10} = 8 \frac {1023}{1}$
$S_{10} = 8 (1023)$
$S_{10} = 8184$

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