What is the sum from n=0 to infinity of (3^n)/(5^n) ?

The sum from n=0 to infinity of (3^n)/(5^n) is converges

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sum from n=0 to infinity of (3^n)/(5^n)

n = 0 \sum \infty \frac{3 ^{n}}{5 ^{n}}

converges

Solution steps

n = 0 \sum \infty \frac{3 ^{n}}{5 ^{n}}

n \to \infty lim (n (\frac{\frac{3 ^{n}}{5 ^{n}}}{\frac{3 ^{n + 1}}{5 ^{n + 1}}} - 1)) = \infty

L > 1,

by the Raabe's test

= converges

y = 6 sin (x), at x = \frac{π}{6}

\frac{\partial}{\partial y} (x y - \frac{x ^{4}}{4})

\int e^{- 2 x} sin (6 x) d x

\frac{d}{d x} (x^{6} e^{2.5 x})

\frac{d}{d x} (e^{- 0.02 x})

What is the sum from n=0 to infinity of (3^n)/(5^n) ?
The sum from n=0 to infinity of (3^n)/(5^n) is converges