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

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

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

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

n = 0 \sum \infty \frac{1}{( n ( n + 1 ) )}

n = 0 \sum \infty 1 + (- 1)^{n}

n = 0 \sum \infty ln (5 - n)

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

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

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