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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Simplify

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

Apply Cauchy's Convergence Condition:

diverges

= - diverges

= diverges

f (x) = (x^{2} 6^{x})^{9}

\frac{d}{d t} (3.5 sin (5 t))

(\frac{1}{x ^{3}})^{^{'}}

y = (3 x)^{5 x}

y = 9 x, y = \frac{3}{4} x, y = 70 - x^{2}

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