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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

2 x y d x = (1 + y) d y

\frac{\partial}{\partial y} (x^{3} - 3 x y^{2} + y^{3})

f (x) = \frac{2 t + 3}{t + 5}

\frac{d ^{2}}{d x ^{2}} (t e^{- t})

\int x - 2 x d x

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