What is the sum from n=1 to infinity of (n^5)/(3^n) ?

The sum from n=1 to infinity of (n^5)/(3^n) is converges

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sum from n=1 to infinity of (n^5)/(3^n)

n = 1 \sum \infty \frac{n ^{5}}{3 ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{n ^{5}}{3 ^{n}}

Apply Series Ratio Test:

converges

= converges

\int sec (x) (tan (x) + cos (x)) d x

3 y^{^{'}} = \frac{3 x \cdot y + 18 x ^{2} + 2 y ^{2}}{x ^{2}}, y (1) = 3

f (x) = x^{4} (3 - x)^{2}, at x = 1

\frac{d y}{d x} = 200 - \frac{3 y}{100 - 10 x}

f (x) = 4 x^{2} + 7 x

What is the sum from n=1 to infinity of (n^5)/(3^n) ?
The sum from n=1 to infinity of (n^5)/(3^n) is converges