What is the sum from n=6 to infinity of (3^n)/(12^n) ?

The sum from n=6 to infinity of (3^n)/(12^n) is converges

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sum from n=6 to infinity of (3^n)/(12^n)

n = 6 \sum \infty \frac{3 ^{n}}{1 2 ^{n}}

converges

Solution steps

n = 6 \sum \infty \frac{3 ^{n}}{1 2 ^{n}}

n \to \infty lim (n (\frac{\frac{3 ^{n}}{1 2 ^{n}}}{\frac{3 ^{n + 1}}{1 2 ^{n + 1}}} - 1)) = \infty

L > 1,

by the Raabe's test

= converges

\int_{0}^{15} t + 1 d t

\frac{d}{d x} ((1 + x^{3} Y^{2} Z)^{2})

\frac{d}{d x} (4^{x} ln (4))

y^{^{'}} (y - x^{2}) = - 2 y

\frac{\partial}{\partial x} (ln (x^{4} + y^{2}))

What is the sum from n=6 to infinity of (3^n)/(12^n) ?
The sum from n=6 to infinity of (3^n)/(12^n) is converges