What is the sum from n=1 to infinity}((3^{n+1 of))/((4^{n-1))} ?

The sum from n=1 to infinity}((3^{n+1 of))/((4^{n-1))} is converges

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sum from n=1 to infinity}((3^{n+1 of))/((4^{n-1))}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

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n = 0 \sum \infty \frac{n !}{3 n ! - 1}

What is the sum from n=1 to infinity}((3^{n+1 of))/((4^{n-1))} ?
The sum from n=1 to infinity}((3^{n+1 of))/((4^{n-1))} is converges