What is the sum from n=1 to infinity of (n^4)/(3^n) ?

The sum from n=1 to infinity of (n^4)/(3^n) is converges

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sum from n=1 to infinity of (n^4)/(3^n)

n = 1 \sum \infty \frac{n ^{4}}{3 ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{n ^{4}}{3 ^{n}}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty \frac{7}{n ^{2} + 9}

n = 1 \sum \infty \frac{9}{n ^{2} + n ^{3}}

k = 3 \sum \infty \frac{7}{k ln ^{2} ( k )}

k = 2 \sum \infty \frac{k}{ln ( 5 k )}

n = 2 \sum \infty (\frac{1}{9 n - 2})^{7}

What is the sum from n=1 to infinity of (n^4)/(3^n) ?
The sum from n=1 to infinity of (n^4)/(3^n) is converges