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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

x \to 1 lim (4 x - 1)

\frac{d}{d x} (\frac{( - ln ( 1 + 6 x ) )}{x})

\frac{150000}{( 4 t + 75 ) ^{2}}

\int \frac{1}{6 x} d x

g (x) = \frac{3}{x ^{3}}

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