What is the sum from n=1 to infinity of 3^n4^{-n+1} ?

The sum from n=1 to infinity of 3^n4^{-n+1} is converges

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sum from n=1 to infinity of 3^n4^{-n+1}

n = 1 \sum \infty 3^{n} 4^{- n + 1}

converges

Solution steps

n = 1 \sum \infty 3^{n} \cdot 4^{- n + 1}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty \frac{9}{1 0 ^{n}}

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

n = 1 \sum \infty (\frac{11}{5})^{n}

k = 1 \sum \infty \frac{1}{1 1 ^{k}}

n = 1 \sum \infty \frac{( 1 - 3 ) ^{n}}{n}

What is the sum from n=1 to infinity of 3^n4^{-n+1} ?
The sum from n=1 to infinity of 3^n4^{-n+1} is converges