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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

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

n = 1 \sum \infty 0. 9^{n}

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

n = 0 \sum \infty \frac{3}{5 ^{n - 1}}

n = 0 \sum \infty n - \frac{π}{2}

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