What is the sum from n=1 to infinity of n^{-5/2} ?

The sum from n=1 to infinity of n^{-5/2} 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 n^{-5/2}

n = 1 \sum \infty n^{- \frac{5}{2}}

converges

Solution steps

n = 1 \sum \infty n^{- \frac{5}{2}}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= n = 1 \sum \infty \frac{1}{n ^{\frac{5}{2}}}

Apply p-Series Test:

converges

= converges

k = 1 \sum \infty sin (\frac{1}{k})

n = 1 \sum \infty \frac{1}{( 3 n + 1 ) ^{2}}

n = 1 \sum \infty cos (\frac{11}{n})

n = 0 \sum \infty 3^{n} \cdot 7^{- n}

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

What is the sum from n=1 to infinity of n^{-5/2} ?
The sum from n=1 to infinity of n^{-5/2} is converges