What is the sum from n=2 to infinity of (-1)/n ?

The sum from n=2 to infinity of (-1)/n is diverges

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sum from n=2 to infinity of (-1)/n

n = 2 \sum \infty \frac{- 1}{n}

diverges

Solution steps

n = 2 \sum \infty \frac{- 1}{n}

Apply the constant multiplication rule:

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

= - n = 2 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= - diverges

= diverges

\frac{d}{d x} (\frac{x ^{3} - 2 x ^{2} + 2 x - 4}{x ^{2} + 2})

f (x) = \frac{c ^{9} - x ^{9}}{c ^{9} + x ^{9}}

\frac{d}{d x} (1 + 36 x)

\int \frac{1}{x ^{2} - 12} d x

\int \frac{x + 4}{x ^{3} + 3 x ^{2} - 10 x} d x

What is the sum from n=2 to infinity of (-1)/n ?
The sum from n=2 to infinity of (-1)/n is diverges