What is the sum from n=0 to infinity of 1/(3n^2+8) ?

The sum from n=0 to infinity of 1/(3n^2+8) is converges

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sum from n=0 to infinity of 1/(3n^2+8)

n = 0 \sum \infty \frac{1}{3 n ^{2} + 8}

converges

Solution steps

n = 0 \sum \infty \frac{1}{3 n ^{2} + 8}

Apply Series Comparison Test:

converges

= converges

n = 0 \sum \infty n \cdot 0. 5^{n}

n = 1 \sum \infty \frac{1}{∣ n ∣}

n = 0 \sum \infty (\frac{- 1}{7})^{n}

n = 1 \sum \infty 8 (0.3)^{n - 1}

n = 0 \sum \infty \frac{10}{n ^{2} + 9}

What is the sum from n=0 to infinity of 1/(3n^2+8) ?
The sum from n=0 to infinity of 1/(3n^2+8) is converges