What is the sum from n=1 to infinity of 5/(n^{9/8)} ?

The sum from n=1 to infinity of 5/(n^{9/8)} is converges

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sum from n=1 to infinity of 5/(n^{9/8)}

n = 1 \sum \infty \frac{5}{n ^{\frac{9}{8}}}

converges

Solution steps

n = 1 \sum \infty \frac{5}{n ^{\frac{9}{8}}}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{n ^{\frac{9}{8}}}

Apply exponent rule:

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

\frac{1}{n ^{\frac{9}{8}}} = n^{- \frac{9}{8}}

= 5 \cdot n = 1 \sum \infty n^{- \frac{9}{8}}

Apply Series Integral Test:

converges

= 5 converges

= converges

n = 0 \sum \infty sin (\frac{1}{n ^{0}})

n = 1 \sum \infty \frac{5}{n + 4}

n = 2 \sum \infty (- 0.95)^{n}

n = 1 \sum \infty 10 (- \frac{3}{8})^{n}

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

What is the sum from n=1 to infinity of 5/(n^{9/8)} ?
The sum from n=1 to infinity of 5/(n^{9/8)} is converges