What is the sum from n=3 to infinity of 1/(n^{100)} ?

The sum from n=3 to infinity of 1/(n^{100)} 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=3 to infinity of 1/(n^{100)}

n = 3 \sum \infty \frac{1}{n ^{100}}

converges

Solution steps

n = 3 \sum \infty \frac{1}{n ^{100}}

Apply p-Series Test:

converges

= converges

n = 0 \sum \infty \frac{3 ^{n}}{1 0 ^{n}}

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

j = 1 \sum \infty 3^{- 4 j}

n = 1 \sum \infty ln (\frac{10}{n})

n = 0 \sum \infty \frac{1}{π ^{n}}

What is the sum from n=3 to infinity of 1/(n^{100)} ?
The sum from n=3 to infinity of 1/(n^{100)} is converges