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

The sum from n=2 to infinity}(e^{1/n of)/n is converges

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sum from n=2 to infinity}(e^{1/n of)/n

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

converges

Solution steps

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

Apply Series Integral Test:

converges

= converges

n = 0 \sum \infty (\frac{4}{3})^{1 - n}

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

n = 1 \sum \infty 10 (- \frac{7}{8})^{n}

n = 0 \sum \infty 4 n

n = 0 \sum \infty 12

What is the sum from n=2 to infinity}(e^{1/n of)/n ?
The sum from n=2 to infinity}(e^{1/n of)/n is converges