What is the sum from n=1 to infinity of 2/(n^{0.85)} ?

The sum from n=1 to infinity of 2/(n^{0.85)} is diverges

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sum from n=1 to infinity of 2/(n^{0.85)}

n = 1 \sum \infty \frac{2}{n ^{0.85}}

diverges

Solution steps

n = 1 \sum \infty \frac{2}{n ^{0.85}}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{1}{n ^{0.85}}

Apply p-Series Test:

diverges

= 2 diverges

= diverges

n = 1 \sum \infty (- \frac{6}{7})^{n - 1}

n = 0 \sum \infty \frac{e ^{n}}{n + 1}

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

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

n = 0 \sum \infty \frac{3 n}{n + 2}

What is the sum from n=1 to infinity of 2/(n^{0.85)} ?
The sum from n=1 to infinity of 2/(n^{0.85)} is diverges