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

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

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

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

diverges

Solution steps

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

Simplify

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

Apply Cauchy's Convergence Condition:

diverges

= diverges

n = 1 \sum \infty \frac{1}{16 n ^{2} - 8 n - 3}

n = 0 \sum \infty \frac{2}{( 1 - n ) ^{3}}

n = 1 \sum \infty \frac{5}{( - 4 ) ^{n}}

n = 1 \sum \infty \frac{n !}{11 4 ^{n}}

n = 7 \sum \infty \frac{1}{n} - \frac{1}{n + 3}

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