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

The sum from n=1 to infinity of (6^n)/(n^k) is diverges

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

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

diverges

Solution steps

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

n \to \infty lim n \frac{\frac{6 ^{n}}{n ^{k}}}{\frac{6 ^{n + 1}}{( n + 1 ) ^{k}}} - 1 = - \infty

L < 1,

by the Raabe's test

= diverges

n = 5 \sum \infty \frac{3}{2 ^{n}}

k = 0 \sum \infty (- \frac{1}{4})^{k}

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

n = 0 \sum \infty \frac{7}{n}

n = 1 \sum \infty n! (- e)^{- 6 n}

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