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

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

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

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

diverges

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}

Simplify

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

Apply Cauchy's Convergence Condition:

diverges

= - diverges

= diverges

n = 0 \sum \infty e^{\frac{1}{2}}

n = 1 \sum \infty \frac{1}{ln ( 5 ^{n} )}

n = 0 \sum \infty (\frac{1}{1.02})^{n}

k = 2 \sum \infty \frac{k}{k ^{3} - k + 1}

n = 2 \sum \infty \frac{2}{n \cdot ( n + 2 )}

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