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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

n = 1 \sum \infty cos (\frac{11}{n})

n = 0 \sum \infty 3^{n} \cdot 7^{- n}

n = 1 \sum \infty \frac{n}{n n}

n = 1 \sum \infty \frac{2}{3 n + 7}

n = 0 \sum \infty (\frac{10 π}{e})^{n}

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