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find the sum of the infinite series: E 3(-1/3)^i i=1 any help is appreciated, thanks!
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\[3 (\frac{ 1 }{ 3 })^{i}\] ?
3(1/3)^i..... i being the variable, could also be n if needs be
Since the number being raised to a power has an absolute value of less than one, we can get a sum from the infinite series. S(infinity) = a1 / (1-r) a1 is the first value of the series, r is the common ratio (repeated multiplier)
swell, thank you!
Do you see? I don't mean to confuse and I can simplify.
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naww its all good I got it, I wasn't sure what formula to use, thanks!
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