What is the sum from n=1 to infinity of 3/(n^9) ?

The sum from n=1 to infinity of 3/(n^9) is converges

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sum from n=1 to infinity of 3/(n^9)

n = 1 \sum \infty \frac{3}{n ^{9}}

converges

Solution steps

n = 1 \sum \infty \frac{3}{n ^{9}}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{1}{n ^{9}}

Apply p-Series Test:

converges

= 3 converges

= converges

n = 0 \sum \infty \frac{n ^{2}}{1 + 6 n}

n = 1 \sum \infty \frac{2}{n ^{9}}

n = 1 \sum \infty (0.75)^{n - 1}

n = 1 \sum \infty \frac{3 ^{n}}{7 ^{n}}

n = 3 \sum \infty \frac{6}{n ^{3}}

What is the sum from n=1 to infinity of 3/(n^9) ?
The sum from n=1 to infinity of 3/(n^9) is converges