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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

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

n = 0 \sum \infty \frac{1}{1 - 6 n}

n = 0 \sum \infty (0.2)^{n}

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

n = 0 \sum \infty n \cdot e^{- n}

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