What is the sum from n=1 to infinity of (-2^n)/(n^7) ?

The sum from n=1 to infinity of (-2^n)/(n^7) 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 of (-2^n)/(n^7)

n = 1 \sum \infty \frac{- 2 ^{n}}{n ^{7}}

diverges

Solution steps

n = 1 \sum \infty \frac{- 2 ^{n}}{n ^{7}}

Apply the constant multiplication rule:

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

= - n = 1 \sum \infty \frac{2 ^{n}}{n ^{7}}

Apply Series Ratio Test:

diverges

= - diverges

= diverges

\frac{d}{d x} (\frac{12 - 3 y - 2 x}{4})

y = x (x^{2} + 1)^{3}

\int (cos (n x)) d x

\int_{- 1}^{2} (1 - x) d x

n = 1 \sum \infty \frac{1}{3 ^{n}}

What is the sum from n=1 to infinity of (-2^n)/(n^7) ?
The sum from n=1 to infinity of (-2^n)/(n^7) is diverges