Monday

December 22, 2014

December 22, 2014

Posted by **SDSU** on Sunday, July 12, 2009 at 11:36pm.

- calculus II -
**drwls**, Monday, July 13, 2009 at 1:37amCalculate the limit of the series

100[1 + 2*(2/3) + 2*(2/3)^2 + 2*(2/3)^3+...]

= 100 + 2[1 + 2/3 + (2/3)^2+ ...]

Make use of the relation:

1 + x + x^2 + .. = 1/(1-x)

(for x<1, which is the case here)

- calculus II (correction) -
**drwls**, Monday, July 13, 2009 at 8:37am= 100 [1 + (4/3)[1 + 2/3 + (2/3)^2+ ...]

= 100 {1 + (4/3)[1/(1 - 2/3)]} = ?

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