What is the sum from n=1 to infinity of 7/(n^{2/5)} ?

The sum from n=1 to infinity of 7/(n^{2/5)} is diverges

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sum from n=1 to infinity of 7/(n^{2/5)}

n = 1 \sum \infty \frac{7}{n ^{\frac{2}{5}}}

diverges

Solution steps

n = 1 \sum \infty \frac{7}{n ^{\frac{2}{5}}}

Apply the constant multiplication rule:

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

= 7 \cdot n = 1 \sum \infty \frac{1}{n ^{\frac{2}{5}}}

Apply p-Series Test:

diverges

= 7 diverges

= diverges

n = 1 \sum \infty \frac{6 n}{3 n + 9}

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

n = 0 \sum \infty cos (\frac{nπ}{3})

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

n = 1 \sum \infty - 8 (\frac{4}{5})^{n - 1}

What is the sum from n=1 to infinity of 7/(n^{2/5)} ?
The sum from n=1 to infinity of 7/(n^{2/5)} is diverges