What is the sum from n=0 to infinity of (-1^n)/n ?

The sum from n=0 to infinity of (-1^n)/n is diverges

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sum from n=0 to infinity of (-1^n)/n

n = 0 \sum \infty \frac{- 1 ^{n}}{n}

diverges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Simplify

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

Apply Cauchy's Convergence Condition:

diverges

= - diverges

= diverges

n = 1 \sum \infty \frac{6}{n ^{\frac{4}{3}}}

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

k = 1 \sum \infty 5 (- \frac{1}{8})^{5 k}

n = 1 \sum \infty 5 (0.3)^{n - 1}

n = 1 \sum \infty cos (5 n)

What is the sum from n=0 to infinity of (-1^n)/n ?
The sum from n=0 to infinity of (-1^n)/n is diverges