anonymous
  • anonymous
Find the sum of the Series: Sigma (-1)^(n-1) (3/4^n) from n=1 to infinity
Mathematics
jamiebookeater
  • jamiebookeater
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blockcolder
  • blockcolder
You mean this: \[\sum_{n=1}^{\infty}(-1)^{n-1}\frac{3}{4^n}\]
anonymous
  • anonymous
Yes
blockcolder
  • blockcolder
This is an alternating geometric series with r=-1/4 and a=3/4. Thus, the sum is \[\frac{\frac{3}{4}}{1+\frac{1}{4}}=\frac{3}{5}\].

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campbell_st
  • campbell_st
the common ratio is -1/4 and 1st term a = 3/4 then \[s _{\infty} = a/(1-r)\]
anonymous
  • anonymous
Thank you!
blockcolder
  • blockcolder
No problem. :D

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