What is the sum from n=0 to infinity of 5^n4^{-n} ?

The sum from n=0 to infinity of 5^n4^{-n} is diverges

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sum from n=0 to infinity of 5^n4^{-n}

n = 0 \sum \infty 5^{n} 4^{- n}

diverges

Solution steps

n = 0 \sum \infty 5^{n} \cdot 4^{- n}

Simplify

4^{- n} : \frac{1}{4 ^{n}}

= n = 0 \sum \infty 5^{n} \cdot \frac{1}{4 ^{n}}

Simplify

5^{n} \cdot \frac{1}{4 ^{n}} : (\frac{5}{4})^{n}

= n = 0 \sum \infty (\frac{5}{4})^{n}

Apply Series Geometric Test:

diverges

= diverges

k = 1 \sum \infty 5^{- k}

n = 1 \sum \infty n (\frac{- 5}{9})^{n}

n = 1 \sum \infty \frac{2}{n ^{2} + 6 n}

n = 2 \sum \infty \frac{1}{n ^{1.1}}

n = 5 \sum \infty \frac{1}{n} - \frac{1}{n + 3}

What is the sum from n=0 to infinity of 5^n4^{-n} ?
The sum from n=0 to infinity of 5^n4^{-n} is diverges