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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty (- \frac{1}{7})^{n}

n = 1 \sum \infty sin (\frac{πn}{2})

n = 1 \sum \infty 0. 9^{n - 1}

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

j = 1 \sum \infty 2^{- 4 j}

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