What is the sum from n=1 to infinity of 1/(9n^2+23) ?

The sum from n=1 to infinity of 1/(9n^2+23) is converges

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sum from n=1 to infinity of 1/(9n^2+23)

n = 1 \sum \infty \frac{1}{9 n ^{2} + 23}

converges

Solution steps

n = 1 \sum \infty \frac{1}{9 n ^{2} + 23}

Apply Series Comparison Test:

converges

= converges

m = 2 \sum \infty \frac{6}{7 ^{m}}

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

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

n = 0 \sum \infty (\frac{- 7}{8})^{n}

k = 1 \sum \infty 1^{k}

What is the sum from n=1 to infinity of 1/(9n^2+23) ?
The sum from n=1 to infinity of 1/(9n^2+23) is converges