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

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

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

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

n = 1 \sum \infty \frac{6}{5 ^{n}}

n = 0 \sum \infty \frac{2 ^{n}}{n ^{4}}

n = 1 \sum \infty (0.7)^{n}

n = 0 \sum \infty (- \frac{1}{3})^{n - 1}

k = 0 \sum \infty (\frac{1}{19})^{k}

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