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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

converges

= 9 converges

= converges

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

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

n = 0 \sum \infty \frac{n}{5 n + 1}

n = 1 \sum \infty \frac{1}{n} sin (\frac{1}{n})

n = 1 \sum \infty \frac{1 0 ^{n}}{n + 1}

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