What is the sum from n=1 to infinity of 1/((9n)^9) ?

The sum from n=1 to infinity of 1/((9n)^9) is converges

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sum from n=1 to infinity of 1/((9n)^9)

n = 1 \sum \infty \frac{1}{( 9 n ) ^{9}}

converges

Solution steps

n = 1 \sum \infty \frac{1}{( 9 n ) ^{9}}

Simplify

\frac{1}{( 9 n ) ^{9}} : \frac{1}{387420489 n ^{9}}

= n = 1 \sum \infty \frac{1}{387420489 n ^{9}}

Apply the constant multiplication rule:

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

= \frac{1}{387420489} \cdot n = 1 \sum \infty \frac{1}{n ^{9}}

Apply p-Series Test:

converges

= \frac{1}{387420489} converges

= converges

k = 2 \sum \infty (- 0.15)^{k}

n = 5 \sum \infty \frac{e ^{n}}{5 ^{n}}

n = 1 \sum \infty (\frac{9}{4})^{n}

n = 2 \sum \infty \frac{1}{ln ( n ^{n} )}

n = 0 \sum \infty 2^{n} 3^{n - 2}

What is the sum from n=1 to infinity of 1/((9n)^9) ?
The sum from n=1 to infinity of 1/((9n)^9) is converges