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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

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

n = 2 \sum \infty (- 0.7)^{n}

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

n = 1 \sum \infty \frac{1}{6 + e ^{- n}}

n = 2 \sum \infty \frac{2}{n ln ( n )}

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