What is the sum from n=1 to infinity of 9/(n^2+1) ?

The sum from n=1 to infinity of 9/(n^2+1) is converges

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sum from n=1 to infinity of 9/(n^2+1)

n = 1 \sum \infty \frac{9}{n ^{2} + 1}

converges

Solution steps

n = 1 \sum \infty \frac{9}{n ^{2} + 1}

Apply the constant multiplication rule:

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

= 9 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + 1}

Apply Series Comparison Test:

converges

= 9 converges

= converges

\int_{1}^{16} \frac{1}{10 + x} d x

x \to 1 + lim (\frac{3 + x}{x ^{2} - 1})

f (x) = \frac{4 x}{x ^{2} + 1}, at x = - 1

\int \frac{x}{x ^{2} + 3 x + 3} d x

What is the sum from n=1 to infinity of 9/(n^2+1) ?
The sum from n=1 to infinity of 9/(n^2+1) is converges