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 converges

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

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

converges

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}

Apply Alternating Series Test:

converges

= - converges

= converges

n = 500 \sum \infty \frac{1}{n}

n = 2 \sum \infty \frac{ln ( n )}{n}

x = 0 \sum \infty \frac{x}{1 - 4 x}

n = 3 \sum \infty \frac{1}{n \cdot ln ( n )}

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

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 converges