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\( \sum _ { n = 1 } ^ { \infty } \frac { x ^ { n } } { 4 ^ { n } } \)

Ask by Riley Guerrero. in the United States
Mar 14,2025

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The sum of the series \( \sum_{n=1}^{\infty} \frac{x^n}{4^n} \) is \( \frac{\frac{x}{4}}{1 - \frac{x}{4}} = \frac{x}{4 - x} \), provided that \( |x| < 4 \).

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Bonus Knowledge

This series can be recognized as a geometric series of the form \( \sum_{n=0}^\infty ar^n \), where the first term \( a = \frac{x}{4} \) and the common ratio \( r = \frac{x}{4} \). For such series, the sum converges when the absolute value of the ratio is less than one, specifically \( |r| < 1 \) or \( |x| < 4 \). When it converges, the sum is given by the formula \( \frac{a}{1 - r} \). Thus, the sum of your series can be expressed as: \[ \sum _ { n = 1 } ^ { \infty } \frac { x ^ { n } } { 4 ^ { n } } = \frac{\frac{x}{4}}{1 - \frac{x}{4}} = \frac{x/4}{(4-x)/4} = \frac{x}{4-x} \] This is true for \( |x| < 4 \). It's a handy formula you can use when dealing with series like this in calculus or series summation problems!

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