What is the sum from n=1 to infinity of (5/n)^4 ?

The sum from n=1 to infinity of (5/n)^4 is converges

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sum from n=1 to infinity of (5/n)^4

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

converges

Solution steps

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

Simplify

(\frac{5}{n})^{4} : \frac{625}{n ^{4}}

= n = 1 \sum \infty \frac{625}{n ^{4}}

Apply the constant multiplication rule:

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

= 625 \cdot n = 1 \sum \infty \frac{1}{n ^{4}}

Apply p-Series Test:

converges

= 625 converges

= converges

d er i v a t i v e f (x) = \frac{x + 7}{8 x}

\frac{d}{d x} (\frac{1}{sin ^{3} ( x )})

t an g e n t y = \frac{7}{x}

\frac{\partial}{\partial x} (x^{2} (x - 2)^{4})

x^{2} y^{^{''}} + 3 x y^{^{'}} - 3 y = 0, y (1) = 0, y^{^{'}} (1) = 5

What is the sum from n=1 to infinity of (5/n)^4 ?
The sum from n=1 to infinity of (5/n)^4 is converges