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

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

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sum from n=1 to infinity of (1^n)/n

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

diverges

Solution steps

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

Simplify

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

Apply Cauchy's Convergence Condition:

diverges

= diverges

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

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

n = 3 \sum \infty \frac{1}{2 n ^{2} + n}

n = 1 \sum \infty \frac{8}{( - 3 ) ^{n}}

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

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