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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

n = 0 \sum \infty \frac{1}{7 ^{n}}

\frac{d y}{d x} = \frac{( x + sec ^{2} ( x ) )}{y}

\int \frac{x}{x ^{4} + 16} d x

a re a y = 3 x^{2} - 12, y = 0

\frac{d}{d x} (\frac{3 + sin ( x )}{3 x + cos ( x )})

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