What is the sum from n=1 to infinity of (8^n)/(n^4) ?

The sum from n=1 to infinity of (8^n)/(n^4) is diverges

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sum from n=1 to infinity of (8^n)/(n^4)

n = 1 \sum \infty \frac{8 ^{n}}{n ^{4}}

diverges

Solution steps

n = 1 \sum \infty \frac{8 ^{n}}{n ^{4}}

n \to \infty lim (n (\frac{\frac{8 ^{n}}{n ^{4}}}{\frac{8 ^{n + 1}}{( n + 1 ) ^{4}}} - 1)) = - \infty

L < 1,

by the Raabe's test

= diverges

\int_{0}^{\infty} \frac{1}{1 + e ^{x}} d x

(x + 1) \frac{d y}{d x} + (x + 2) y = 4 x e^{- x}

\int \frac{3}{2 t - 5} d t

\int \frac{x}{16 - x ^{4}} d x

\frac{\partial}{\partial x} (x - x y + 3)

What is the sum from n=1 to infinity of (8^n)/(n^4) ?
The sum from n=1 to infinity of (8^n)/(n^4) is diverges