What is the sum from n=1 to infinity of 5/(n^2+49) ?

The sum from n=1 to infinity of 5/(n^2+49) is converges

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sum from n=1 to infinity of 5/(n^2+49)

n = 1 \sum \infty \frac{5}{n ^{2} + 49}

converges

Solution steps

n = 1 \sum \infty \frac{5}{n ^{2} + 49}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + 49}

Apply Series Comparison Test:

converges

= 5 converges

= converges

n = 0 \sum \infty \frac{n}{n ^{2} - 2 n + 1}

n = 0 \sum \infty \frac{7}{n} - \frac{7}{n + 1}

n = 1 \sum \infty (\frac{4}{5})^{n + 1}

n = 0 \sum \infty \frac{1}{2 + n}

n = 0 \sum \infty \frac{1}{( 1 - n ) ^{3}}

What is the sum from n=1 to infinity of 5/(n^2+49) ?
The sum from n=1 to infinity of 5/(n^2+49) is converges