What is the sum from n=1 to infinity of 6/(n^{4/3)} ?

The sum from n=1 to infinity of 6/(n^{4/3)} is converges

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sum from n=1 to infinity of 6/(n^{4/3)}

n = 1 \sum \infty \frac{6}{n ^{\frac{4}{3}}}

converges

Solution steps

n = 1 \sum \infty \frac{6}{n ^{\frac{4}{3}}}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{1}{n ^{\frac{4}{3}}}

Apply p-Series Test:

converges

= 6 converges

= converges

n = 0 \sum \infty \frac{n !}{n ! + 1}

k = 1 \sum \infty 5 (- \frac{1}{8})^{5 k}

n = 1 \sum \infty 5 (0.3)^{n - 1}

n = 1 \sum \infty cos (5 n)

n = 0 \sum \infty \frac{1}{7 - n}

What is the sum from n=1 to infinity of 6/(n^{4/3)} ?
The sum from n=1 to infinity of 6/(n^{4/3)} is converges