What is the sum from n=5 to infinity of (8^n)/(9^n) ?

The sum from n=5 to infinity of (8^n)/(9^n) is converges

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sum from n=5 to infinity of (8^n)/(9^n)

n = 5 \sum \infty \frac{8 ^{n}}{9 ^{n}}

converges

Solution steps

n = 5 \sum \infty \frac{8 ^{n}}{9 ^{n}}

n \to \infty lim (n (\frac{\frac{8 ^{n}}{9 ^{n}}}{\frac{8 ^{n + 1}}{9 ^{n + 1}}} - 1)) = \infty

L > 1,

by the Raabe's test

= converges

\int \frac{3}{5} - \frac{4}{x} d x

n = 1 \sum \infty n (sin (\frac{1}{n}))

\int (4 t^{2} - 3)^{2} d t

f (x) = x^{2}, [1, 5]

\frac{8 x ^{2} + 2}{x ^{2}}

What is the sum from n=5 to infinity of (8^n)/(9^n) ?
The sum from n=5 to infinity of (8^n)/(9^n) is converges