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

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

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

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

converges

Solution steps

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

Simplify

(\frac{4}{n})^{4} : \frac{256}{n ^{4}}

= n = 1 \sum \infty \frac{256}{n ^{4}}

Apply the constant multiplication rule:

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

= 256 \cdot n = 1 \sum \infty \frac{1}{n ^{4}}

Apply p-Series Test:

converges

= 256 converges

= converges

n = 1 \sum \infty \frac{7}{n !}

n = 0 \sum \infty (\frac{7}{6})^{n}

n = 1 \sum \infty \frac{1}{3 n ^{2}}

k = 0 \sum \infty (\frac{1}{4})^{k}

n = 0 \sum \infty \frac{5}{n}

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