What is the sum from n=6 to infinity of (5^n)/(12^n) ?

The sum from n=6 to infinity of (5^n)/(12^n) is converges

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sum from n=6 to infinity of (5^n)/(12^n)

n = 6 \sum \infty \frac{5 ^{n}}{1 2 ^{n}}

converges

Solution steps

n = 6 \sum \infty \frac{5 ^{n}}{1 2 ^{n}}

n \to \infty lim (n (\frac{\frac{5 ^{n}}{1 2 ^{n}}}{\frac{5 ^{n + 1}}{1 2 ^{n + 1}}} - 1)) = \infty

L > 1,

by the Raabe's test

= converges

3 y^{^{''}} + 17 y^{^{'}} - 6 y = 0

\frac{d}{d x} (\frac{1}{x ^{4}} + \frac{1}{x ^{2}} - x^{3})

\int (x^{2} - 5)^{5} \cdot 2 x d x

y^{^{''}} = x e^{x}

f (x) = x^{3} - 2 x^{2} + 4 x + 3, at x = 1

What is the sum from n=6 to infinity of (5^n)/(12^n) ?
The sum from n=6 to infinity of (5^n)/(12^n) is converges