What is the sum from n=1 to infinity of 5/(1+n^2) ?

The sum from n=1 to infinity of 5/(1+n^2) is converges

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sum from n=1 to infinity of 5/(1+n^2)

n = 1 \sum \infty \frac{5}{1 + n ^{2}}

converges

Solution steps

n = 1 \sum \infty \frac{5}{1 + n ^{2}}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{1 + n ^{2}}

Apply Series Comparison Test:

converges

= 5 converges

= converges

x = 0 \sum \infty \frac{1}{1 - x}

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

n = 1 \sum \infty 2 (0.1)^{n - 1}

n = 1 \sum \infty \frac{3}{( 2 n + 5 ) ^{3}}

n = 1 \sum \infty \frac{3 n}{n ^{2} - 7}

What is the sum from n=1 to infinity of 5/(1+n^2) ?
The sum from n=1 to infinity of 5/(1+n^2) is converges