What is the sum from n=5 to infinity of (3^n)/(12^n) ?

The sum from n=5 to infinity of (3^n)/(12^n) is converges

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sum from n=5 to infinity of (3^n)/(12^n)

n = 5 \sum \infty \frac{3 ^{n}}{1 2 ^{n}}

converges

Solution steps

n = 5 \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

t an g e n t y = x^{2} - 5

x \to - 1 lim (4 x^{2} - 2 x - 6)

d er i v a t i v e sin (2 π t)

\frac{d}{d y} (\frac{350}{x ^{2} + y ^{2}})

\frac{\partial}{\partial x} (6 - 2 x^{2} + 3 y^{2})

What is the sum from n=5 to infinity of (3^n)/(12^n) ?
The sum from n=5 to infinity of (3^n)/(12^n) is converges