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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

2 x y d x + (x^{2} + 1) d y = 0

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

f (x) = 3 - 2 x

\frac{d}{d x} (sin (x^{3} - 3 x))

x^{- 1} - \frac{1}{x}

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