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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

diverges

Solution steps

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

Simplify

(- 1)^{2 n} \frac{1}{n} : \frac{1}{n}

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

Apply Cauchy's Convergence Condition:

diverges

= diverges

n = 1 \sum \infty \frac{n ^{2}}{9 ^{n}}

k = 2 \sum \infty \frac{k ^{e}}{k ^{π}}

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

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

n = 0 \sum \infty \frac{6}{1 - n ^{2}}

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