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

The sum from n=1 to infinity of n^{-5} is converges

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

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

converges

Solution steps

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

Apply exponent rule:

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

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

Apply p-Series Test:

converges

= converges

n = 3 \sum \infty \frac{1}{n ^{2} - 4}

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

n = 1 \sum \infty (\frac{2}{13})^{n}

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

n = 0 \sum \infty (4 + n^{2})^{- 1}

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