What is the sum from n=0 to infinity of (3^n)/n ?

The sum from n=0 to infinity of (3^n)/n is diverges

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sum from n=0 to infinity of (3^n)/n

n = 0 \sum \infty \frac{3 ^{n}}{n}

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

\int_{0}^{2} - 3 x^{3} x^{2} + 4 d x

\frac{16}{s ^{2} + 3 s + 16}

\int_{0}^{2} \frac{x}{x ^{2} + 1} d x

(\frac{y}{2} + 2 x) d x + (- 1) d y = 0

y = 3, y = x, x = 0

What is the sum from n=0 to infinity of (3^n)/n ?
The sum from n=0 to infinity of (3^n)/n is diverges