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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Simplify

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

Apply Cauchy's Convergence Condition:

diverges

= - diverges

= diverges

n = 2 \sum \infty \frac{n}{n ^{2} - 2}

n = 0 \sum \infty (\frac{5}{2})^{n - 1}

k = 0 \sum \infty (0.4)^{k}

n = 1 \sum \infty n^{3} \frac{1}{n !}

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

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