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

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

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

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

converges

Solution steps

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

Apply Alternating Series Test:

converges

= converges

\int e^{6 x} d x

\int sin^{2} (ln (x)) d x

\int \frac{2 x - 9}{x ^{2} - 9 x + 1} d x

\frac{\partial}{\partial x} (\frac{x y}{( x ^{2} - y ^{2} ) ^{2}})

y = x^{2} + 1, (- 2, 5)

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