What is the sum from n=1 to infinity of (3/n)^4 ?

The sum from n=1 to infinity of (3/n)^4 is converges

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sum from n=1 to infinity of (3/n)^4

n = 1 \sum \infty (\frac{3}{n})^{4}

converges

Solution steps

n = 1 \sum \infty (\frac{3}{n})^{4}

Simplify

(\frac{3}{n})^{4} : \frac{81}{n ^{4}}

= n = 1 \sum \infty \frac{81}{n ^{4}}

Apply the constant multiplication rule:

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

= 81 \cdot n = 1 \sum \infty \frac{1}{n ^{4}}

Apply p-Series Test:

converges

= 81 converges

= converges

\int \frac{1}{sin ^{2} ( t )} d t

r = 1 0^{2 t}

\frac{\partial}{\partial x} (sin (π (4 x - 3 y)))

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

\frac{\partial}{\partial x} (x + y + ln (x y))

What is the sum from n=1 to infinity of (3/n)^4 ?
The sum from n=1 to infinity of (3/n)^4 is converges