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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

n = 1 \sum \infty \frac{32}{n !}

n = 0 \sum \infty \frac{n}{n ^{2} + 7}

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

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

n = 1 \sum \infty \frac{10}{2 ^{n}}

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