What is the sum from n=1 to infinity of 3/(n^{5/2)} ?

The sum from n=1 to infinity of 3/(n^{5/2)} is converges

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sum from n=1 to infinity of 3/(n^{5/2)}

n = 1 \sum \infty \frac{3}{n ^{\frac{5}{2}}}

converges

Solution steps

n = 1 \sum \infty \frac{3}{n ^{\frac{5}{2}}}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{1}{n ^{\frac{5}{2}}}

Apply p-Series Test:

converges

= 3 converges

= converges

k = 1 \sum \infty k^{2} e^{- k}

n = 1 \sum \infty n! (- e)^{- 4 n}

n = 1 \sum \infty \frac{6}{n ^{2} + 1}

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

n = 1 \sum \infty \frac{19}{1 + 3 ^{n}}

What is the sum from n=1 to infinity of 3/(n^{5/2)} ?
The sum from n=1 to infinity of 3/(n^{5/2)} is converges