What is the sum from n=1 to infinity of 4^n5^{-n} ?

The sum from n=1 to infinity of 4^n5^{-n} is converges

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sum from n=1 to infinity of 4^n5^{-n}

n = 1 \sum \infty 4^{n} 5^{- n}

converges

Solution steps

n = 1 \sum \infty 4^{n} \cdot 5^{- n}

Apply Series Ratio Test:

converges

= converges

x = 1 \sum \infty \frac{2}{x ^{2} + 4 x + 3}

n = 0 \sum \infty \frac{1}{cos ( \frac{1}{n} )}

n = 1 \sum \infty \frac{1}{4 n ^{2} - 81}

n = 0 \sum \infty \frac{1}{n ^{\frac{1}{6}}}

n = 0 \sum \infty 5 (- \frac{1}{2})^{n}

What is the sum from n=1 to infinity of 4^n5^{-n} ?
The sum from n=1 to infinity of 4^n5^{-n} is converges