What is the sum from n=1 to infinity of 7/(1+n^2) ?

The sum from n=1 to infinity of 7/(1+n^2) is converges

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sum from n=1 to infinity of 7/(1+n^2)

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

= 7 \cdot n = 1 \sum \infty \frac{1}{1 + n ^{2}}

Apply Series Comparison Test:

converges

= 7 converges

= converges

n = 0 \sum \infty 3

n = 1 \sum \infty \frac{2 n}{7 n - 9}

n = 1 \sum \infty \frac{2 n}{7 n - 5}

n = 1 \sum \infty \frac{1}{n ^{\frac{11}{2}}}

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

What is the sum from n=1 to infinity of 7/(1+n^2) ?
The sum from n=1 to infinity of 7/(1+n^2) is converges