What is the sum from n=0 to infinity of n^{-3/2} ?

The sum from n=0 to infinity of n^{-3/2} is converges

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sum from n=0 to infinity of n^{-3/2}

n = 0 \sum \infty n^{- \frac{3}{2}}

converges

Solution steps

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

Apply exponent rule:

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

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

Apply p-Series Test:

converges

= converges

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

n = 1 \sum \infty \frac{9}{7 ^{n}}

n = 5 \sum \infty \frac{4}{n ^{2} - n}

n = 2 \sum \infty \frac{3}{n ^{3}}

n = 7 \sum \infty \frac{1}{n} - \frac{1}{n + 1}

What is the sum from n=0 to infinity of n^{-3/2} ?
The sum from n=0 to infinity of n^{-3/2} is converges