What is the sum from n=1 to infinity of (n^4)/(6^n) ?

The sum from n=1 to infinity of (n^4)/(6^n) is converges

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sum from n=1 to infinity of (n^4)/(6^n)

n = 1 \sum \infty \frac{n ^{4}}{6 ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{n ^{4}}{6 ^{n}}

Apply Series Ratio Test:

converges

= converges

k = 0 \sum \infty (- \frac{1}{9})^{k}

n = 1 \sum \infty \frac{6}{n ^{2}}

x = 2 \sum \infty \frac{1}{x} - \frac{1}{x - 1}

k = 2 \sum \infty \frac{1}{( k - 1 ) ^{4}}

n = 1 \sum \infty \frac{1 ^{n}}{2 n - 1}

What is the sum from n=1 to infinity of (n^4)/(6^n) ?
The sum from n=1 to infinity of (n^4)/(6^n) is converges