What is the sum from n=1 to infinity of 4/(n^2+1) ?

The sum from n=1 to infinity of 4/(n^2+1) is converges

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sum from n=1 to infinity of 4/(n^2+1)

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

converges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply Series Comparison Test:

converges

= 4 converges

= converges

\int - 9.8 d x

\int 2 e^{- 0.2 x} d x

\frac{d ^{2}}{d x ^{2}} y = (4 x + 3) (2 x^{2} + 7 x - 1)

\frac{\partial}{\partial x} (11 x^{2} y + 8 x y^{3})

\frac{d}{d x} (4 - x + 7)

What is the sum from n=1 to infinity of 4/(n^2+1) ?
The sum from n=1 to infinity of 4/(n^2+1) is converges