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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Simplify

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

Apply Cauchy's Convergence Condition:

diverges

= - diverges

= diverges

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