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

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

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

n = 1 \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

\int sin (x) cos (d) x d x

\frac{d y}{d x} = 11 e^{x - y}

\frac{d}{d x} (tan (x) - x)

y^{^{''}} - 2 y^{^{'}} - 3 y = x

\frac{d y}{d t} = t y

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