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

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

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

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

converges

Solution steps

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

Simplify

\frac{1}{( 4 n ) ^{5}} : \frac{1}{1024 n ^{5}}

= n = 1 \sum \infty \frac{1}{1024 n ^{5}}

Apply the constant multiplication rule:

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

= \frac{1}{1024} \cdot n = 1 \sum \infty \frac{1}{n ^{5}}

Apply p-Series Test:

converges

= \frac{1}{1024} converges

= converges

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

n = 1 \sum \infty \frac{ln ( 8 n )}{n}

k = 0 \sum \infty 9 \cdot (\frac{3}{5})^{k}

n = 2 \sum \infty (0.2)^{n}

n = 0 \sum \infty cos (4 n)

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