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Silenthill

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

  • one year ago
  • one year ago

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

    • one year ago
  2. Silenthill
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    Yes

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

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

    • one year ago
  5. Silenthill
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    Thank you!

    • one year ago
  6. blockcolder
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    No problem. :D

    • one year ago
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