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Silenthill

  • 4 years ago

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

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

  2. Silenthill
    • 4 years ago
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    Yes

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

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

  5. Silenthill
    • 4 years ago
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    Thank you!

  6. blockcolder
    • 4 years ago
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    No problem. :D

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