What is the sum from n=0 to infinity of ((-1))/n ?

The sum from n=0 to infinity of ((-1))/n is diverges

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sum from n=0 to infinity of ((-1))/n

n = 0 \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

k = 0 \sum \infty (- \frac{2}{3})^{k}

n = 0 \sum \infty \frac{3}{5 ^{n + 2}}

n = 1 \sum \infty \frac{e}{π ^{n}}

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

n = 1 \sum \infty n e^{2 n}

What is the sum from n=0 to infinity of ((-1))/n ?
The sum from n=0 to infinity of ((-1))/n is diverges