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

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

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

n = 1 \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

\int_{0}^{2} \frac{x ^{2} - 3}{x + 1} d x

\frac{d}{d x} (\frac{ln ( x )}{2 + ln ( x )})

x \to - 3 lim (2 x^{2})

\frac{d}{d x} (\frac{8}{x ^{2} - 4})

\frac{d}{d t} (t sin (2 t))

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