What is the sum from n=6 to infinity of (7^n)/(14^n) ?

The sum from n=6 to infinity of (7^n)/(14^n) is converges

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sum from n=6 to infinity of (7^n)/(14^n)

n = 6 \sum \infty \frac{7 ^{n}}{1 4 ^{n}}

converges

Solution steps

n = 6 \sum \infty \frac{7 ^{n}}{1 4 ^{n}}

n \to \infty lim (n (\frac{\frac{7 ^{n}}{1 4 ^{n}}}{\frac{7 ^{n + 1}}{1 4 ^{n + 1}}} - 1)) = \infty

L > 1,

by the Raabe's test

= converges

\int_{0}^{2} 3 x + 5 d x

\int \frac{1}{x ^{4} + 4} d x

y = \frac{x ^{2} - 2 x}{x}

\frac{\partial}{\partial v} (2 v)

\frac{d y}{d x}, y = - 4 x^{5}

What is the sum from n=6 to infinity of (7^n)/(14^n) ?
The sum from n=6 to infinity of (7^n)/(14^n) is converges