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

\frac{s}{( s ^{2} + 3 ^{2} ) ^{2}}

n = 2 \sum \infty \frac{1}{4 ^{n}}

\int_{0}^{1} t 9 t^{2} + 4 d t

\frac{x}{x + 8}

\int \frac{cos ( x )}{x} d x

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