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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Cauchy's Convergence Condition:

diverges

= - diverges

= diverges

n = 1 \sum \infty \frac{6}{n ^{2} + 3}

n = 1 \sum \infty \frac{1}{n ^{\frac{9}{2}}}

n = 0 \sum \infty (- 0.4)^{n}

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

n = 1 \sum \infty 2 (0.9)^{n - 1}

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