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

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^{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

\int x^{2} cos (\frac{1}{2} x) d x

\int_{0}^{1} \frac{6}{1 + t ^{2}} d t

\frac{d}{d x} (\frac{cos ( x )}{x} + \frac{x}{cos ( x )})

\frac{1}{4 x ^{2} + 1}

x \to 3 - lim (e^{\frac{3}{3 - x}})

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