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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

\frac{d y}{d x}, y = 7^{x}

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

(3 x - 2)^{^{'}}

x \to 3 + lim (\frac{5 x}{x - 3})

f (x) = \frac{6}{3 x + 1}, at x = - 1

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