What is the sum from n=1 to infinity of 3/((4n+2)^3) ?

The sum from n=1 to infinity of 3/((4n+2)^3) is converges

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sum from n=1 to infinity of 3/((4n+2)^3)

n = 1 \sum \infty \frac{3}{( 4 n + 2 ) ^{3}}

converges

Solution steps

n = 1 \sum \infty \frac{3}{( 4 n + 2 ) ^{3}}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{1}{( 4 n + 2 ) ^{3}}

Apply Series Comparison Test:

converges

= 3 converges

= converges

k = 0 \sum \infty 2^{k}

n = 1 \sum \infty \frac{ln ( 5 n )}{n}

k = 2 \sum \infty \frac{1}{k ^{2} - k}

k = 1 \sum \infty \frac{1}{2 ^{k} + k}

k = 3 \sum \infty \frac{1}{k ^{2} - k}

What is the sum from n=1 to infinity of 3/((4n+2)^3) ?
The sum from n=1 to infinity of 3/((4n+2)^3) is converges