What is the sum from n=1 to infinity}((-1)^{2n of)/n ?

The sum from n=1 to infinity}((-1)^{2n of)/n is diverges

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sum from n=1 to infinity}((-1)^{2n of)/n

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

diverges

Solution steps

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

(- 1)^{2 n} = 1

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

Apply Cauchy's Convergence Condition:

diverges

= diverges

n = 1 \sum \infty 2 (\frac{3}{4})^{n - 1}

n = 1 \sum \infty (sin (100))^{n}

n = 1 \sum \infty (\frac{3}{10})^{n}

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

k = 1 \sum \infty \frac{k ^{k}}{k !}

What is the sum from n=1 to infinity}((-1)^{2n of)/n ?
The sum from n=1 to infinity}((-1)^{2n of)/n is diverges