What is the sum from n=3 to infinity of (4^n)/(5^n) ?

The sum from n=3 to infinity of (4^n)/(5^n) is converges

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sum from n=3 to infinity of (4^n)/(5^n)

n = 3 \sum \infty \frac{4 ^{n}}{5 ^{n}}

converges

Solution steps

n = 3 \sum \infty \frac{4 ^{n}}{5 ^{n}}

n \to \infty lim (n (\frac{\frac{4 ^{n}}{5 ^{n}}}{\frac{4 ^{n + 1}}{5 ^{n + 1}}} - 1)) = \infty

L > 1,

by the Raabe's test

= converges

\frac{d}{d x} (a b^{x})

x^{3} (x - 1)^{2}

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

\int \frac{tan ( x )}{1} d x

y^{^{'}} = t - 2 y

What is the sum from n=3 to infinity of (4^n)/(5^n) ?
The sum from n=3 to infinity of (4^n)/(5^n) is converges