What is the sum from n=0 to infinity of 1/((1-3n)^3) ?

The sum from n=0 to infinity of 1/((1-3n)^3) is converges

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sum from n=0 to infinity of 1/((1-3n)^3)

n = 0 \sum \infty \frac{1}{( 1 - 3 n ) ^{3}}

converges

Solution steps

n = 0 \sum \infty \frac{1}{( 1 - 3 n ) ^{3}}

Take the partial fraction of

\frac{1}{( 1 - 3 n ) ^{3}} : - \frac{1}{( 3 n - 1 ) ^{3}}

= n = 0 \sum \infty - \frac{1}{( 3 n - 1 ) ^{3}}

Apply the constant multiplication rule:

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

= - n = 0 \sum \infty \frac{1}{( 3 n - 1 ) ^{3}}

Apply Series Integral Test:

converges

= - converges

= converges

n = 1 \sum \infty (\frac{e}{π})^{n + 1}

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

k = 1 \sum \infty \frac{k ^{6}}{k !}

n = 7 \sum \infty \frac{1}{( 2 n - 1 ) ^{2}}

n = 1 \sum \infty 3^{n} + n + 1

What is the sum from n=0 to infinity of 1/((1-3n)^3) ?
The sum from n=0 to infinity of 1/((1-3n)^3) is converges