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

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

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

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

n = 1 \sum \infty 4 - 9 (0.6)^{n}

n = 2 \sum \infty \frac{6}{n ( n - 1 )}

n = 0 \sum \infty (e)^{- n}

n = 0 \sum \infty (\frac{4}{25})^{n}

n = 2 \sum \infty (\frac{3}{4})^{2 n}

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