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

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

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^7)

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply p-Series Test:

converges

= 2 converges

= converges

n = 1 \sum \infty \frac{1}{n ^{25}}

n = 1 \sum \infty \frac{n !}{3 n ! + 1}

n = 1 \sum \infty 4 (- \frac{10}{11})^{n}

k = 1 \sum \infty \frac{k}{e ^{4 k}}

n = 0 \sum \infty (\frac{4}{3})^{n}

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