What is the sum from n=1 to infinity of 1/n (3/2)^n ?

The sum from n=1 to infinity of 1/n (3/2)^n is diverges

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sum from n=1 to infinity of 1/n (3/2)^n

n = 1 \sum \infty \frac{1}{n} (\frac{3}{2})^{n}

diverges

Solution steps

n = 1 \sum \infty \frac{1}{n} (\frac{3}{2})^{n}

Simplify

\frac{1}{n} (\frac{3}{2})^{n} : \frac{3 ^{n}}{2 ^{n} n}

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

Apply Series Ratio Test:

diverges

= diverges

What is the sum from n=1 to infinity of 1/n (3/2)^n ?
The sum from n=1 to infinity of 1/n (3/2)^n is diverges