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

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

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

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

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

k = 0 \sum \infty (- \frac{7}{8})^{k}

n = 1 \sum \infty \frac{1}{2 n ^{3} + 8}

n = 1 \sum \infty \frac{5}{2 n}

n = 2 \sum \infty \frac{2}{n}

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