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

The sum from n=1 to infinity}(e^{3/n of)/n is converges

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sum from n=1 to infinity}(e^{3/n of)/n

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

converges

Solution steps

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

Apply Series Integral Test:

converges

= converges

n = 0 \sum \infty \frac{1}{7 + n}

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

n = 1 \sum \infty \frac{6}{n + 3}

n = 2 \sum \infty (- 0.65)^{n}

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

What is the sum from n=1 to infinity}(e^{3/n of)/n ?
The sum from n=1 to infinity}(e^{3/n of)/n is converges