What is the sum from n=1 to infinity of (9^n)/(n^5) ?

The sum from n=1 to infinity of (9^n)/(n^5) is diverges

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sum from n=1 to infinity of (9^n)/(n^5)

n = 1 \sum \infty \frac{9 ^{n}}{n ^{5}}

diverges

Solution steps

n = 1 \sum \infty \frac{9 ^{n}}{n ^{5}}

Apply Series Ratio Test:

diverges

= diverges

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

n = 0 \sum \infty 5 (\frac{7}{8})^{n}

n = 0 \sum \infty \frac{1}{n ^{6} + 1}

n = 0 \sum \infty \frac{3 n}{n !}

n = 1 \sum \infty \frac{1}{( n ln ( n ) )}

What is the sum from n=1 to infinity of (9^n)/(n^5) ?
The sum from n=1 to infinity of (9^n)/(n^5) is diverges