What is the sum from n=5 to infinity of (9^n)/(10^n) ?

The sum from n=5 to infinity of (9^n)/(10^n) is converges

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sum from n=5 to infinity of (9^n)/(10^n)

n = 5 \sum \infty \frac{9 ^{n}}{1 0 ^{n}}

converges

Solution steps

n = 5 \sum \infty \frac{9 ^{n}}{1 0 ^{n}}

n \to \infty lim (n (\frac{\frac{9 ^{n}}{1 0 ^{n}}}{\frac{9 ^{n + 1}}{1 0 ^{n + 1}}} - 1)) = \infty

L > 1,

by the Raabe's test

= converges

\frac{d}{d x} (sin (x) - \frac{1}{3} sin^{3} (x))

\int - 25 sin (5 t - \frac{π}{2}) d t

y^{^{'}} + x e^{y} = 0

\int e^{x} 5 + e^{x} d x

f (x) = 3 x sin (x)

What is the sum from n=5 to infinity of (9^n)/(10^n) ?
The sum from n=5 to infinity of (9^n)/(10^n) is converges