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

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

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

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

diverges

Solution steps

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

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

L < 1,

by the Raabe's test

= diverges

\frac{d x}{d t} = - 3 x (\frac{1}{10} e^{- 2 t}), x (0) = \frac{4}{5}

\frac{\partial}{\partial y} (- 3 x^{2} + 7 x y - 4 y^{2} + x + y)

x y^{2} y^{^{'}} = y^{3} - 5 x^{3}, y (1) = 6

\int \frac{1}{2} e^{2 t} d t

\frac{\partial}{\partial z} (5 x^{\frac{y}{z}})

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