How do you find the sum of the series of 3^(k-1)/4^(3k+1) ; k = 0 to infinity? (infinite series)

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How do you find the sum of the series of 3^(k-1)/4^(3k+1) ; k = 0 to infinity? (infinite series)

Mathematics
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You just have to do some manipulation on the series to see that it's actually geometric, and then all you need to do is use the sum of a geometric series formula. So...
\[\sum_{k=0}^{\infty}\frac{3^{k-1}}{4^{3k+1}}=\sum_{k=0}^{\infty}\frac{3^k.3^{-1}}{(4^3)^k.4}=\frac{1}{12}\sum_{k=0}^{k=\infty}(\frac{3}{64})^k\]
The geometric series has first term a=1 and common ratio r=3/64. The sum is then,\[\frac{a}{1-r}=\frac{1}{1-3/64}=\frac{64}{61}\]

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You have to now multiply this answer by 1/12 to get your sum:\[\sum_{k=0}^{\infty}\frac{3^{k-1}}{4^{3k+1}}=\frac{16}{183}\]
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