What is the sum from n=1 to infinity of 6/((3n+1)^2) ?

The sum from n=1 to infinity of 6/((3n+1)^2) is converges

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sum from n=1 to infinity of 6/((3n+1)^2)

n = 1 \sum \infty \frac{6}{( 3 n + 1 ) ^{2}}

converges

Solution steps

n = 1 \sum \infty \frac{6}{( 3 n + 1 ) ^{2}}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{1}{( 3 n + 1 ) ^{2}}

Apply Series Comparison Test:

converges

= 6 converges

= converges

y^{^{''}} = e^{x}

\int_{0}^{16} x x d x

f (x) = ln (x^{2} - 4 x + 1)

y^{^{'}} + 5 x y = 6 x

y = 6 + 5 x^{2} + 2 x^{3}

What is the sum from n=1 to infinity of 6/((3n+1)^2) ?
The sum from n=1 to infinity of 6/((3n+1)^2) is converges