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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

n = 1 \sum \infty \frac{7}{n ^{9}}

n = 0 \sum \infty 1. 1^{n}

n = 1 \sum \infty \frac{3 ^{n}}{n ^{7}}

n = 0 \sum \infty \frac{9}{π ^{n}}

n = 0 \sum \infty \frac{1}{\frac{1}{n}}

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