What is the sum from n=0 to infinity of 1/(4+n^2) ?

The sum from n=0 to infinity of 1/(4+n^2) is converges

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sum from n=0 to infinity of 1/(4+n^2)

n = 0 \sum \infty \frac{1}{4 + n ^{2}}

converges

Solution steps

n = 0 \sum \infty \frac{1}{4 + n ^{2}}

Apply Series Comparison Test:

converges

= converges

n = 1 \sum \infty \frac{ln ( 7 n )}{n}

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

n = 0 \sum \infty 2^{n} (\frac{1}{2})^{n}

n = 1 \sum \infty \frac{n !}{( 4 n ) !}

n = 1 \sum \infty (\frac{7}{5})^{n}

What is the sum from n=0 to infinity of 1/(4+n^2) ?
The sum from n=0 to infinity of 1/(4+n^2) is converges