Silenthill
Find the sum of the Series:
Sigma (1)^(n1) (3/4^n) from n=1 to infinity



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blockcolder
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You mean this:
\[\sum_{n=1}^{\infty}(1)^{n1}\frac{3}{4^n}\]

Silenthill
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Yes

blockcolder
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This is an alternating geometric series with r=1/4 and a=3/4.
Thus, the sum is
\[\frac{\frac{3}{4}}{1+\frac{1}{4}}=\frac{3}{5}\].

campbell_st
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the common ratio is 1/4 and 1st term a = 3/4
then \[s _{\infty} = a/(1r)\]

Silenthill
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Thank you!

blockcolder
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No problem. :D