What is the sum from n=1 to infinity of 3/(1+n^2) ?

The sum from n=1 to infinity of 3/(1+n^2) is converges

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sum from n=1 to infinity of 3/(1+n^2)

n = 1 \sum \infty \frac{3}{1 + n ^{2}}

converges

Solution steps

n = 1 \sum \infty \frac{3}{1 + n ^{2}}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{1}{1 + n ^{2}}

Apply Series Comparison Test:

converges

= 3 converges

= converges

n = 1 \sum \infty 2^{(- 1)^{n}}

n = 1 \sum \infty cos (\frac{17}{n})

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

n = 2 \sum \infty \frac{n !}{n ^{n}}

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

What is the sum from n=1 to infinity of 3/(1+n^2) ?
The sum from n=1 to infinity of 3/(1+n^2) is converges