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

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

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

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

diverges

Solution steps

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

n \to \infty lim (n (\frac{( \frac{3}{2} ) ^{n}}{( \frac{3}{2} ) ^{n + 1}} - 1)) = - \infty

L < 1,

by the Raabe's test

= diverges

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