What is the sum from n=1 to infinity of (8^n)/n ?

The sum from n=1 to infinity of (8^n)/n is diverges

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sum from n=1 to infinity of (8^n)/n

n = 1 \sum \infty \frac{8 ^{n}}{n}

diverges

Solution steps

n = 1 \sum \infty \frac{8 ^{n}}{n}

Apply Series Ratio Test:

diverges

= diverges

n = 2 \sum \infty (\frac{5}{6})^{2 n}

k = 5 \sum \infty \frac{1}{k ^{2} - k}

n = 0 \sum \infty (2)^{n}

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

k = 1 \sum \infty ln (\frac{k}{3 k + 1})

What is the sum from n=1 to infinity of (8^n)/n ?
The sum from n=1 to infinity of (8^n)/n is diverges