What is the sum from n=1 to infinity of 1/(n^2+4n) ?

The sum from n=1 to infinity of 1/(n^2+4n) is 25/48

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sum from n=1 to infinity of 1/(n^2+4n)

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

\frac{25}{48}

Solution steps

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

Take the partial fraction of

\frac{1}{n ^{2} + 4 n} : \frac{1}{4 n} - \frac{1}{4 ( n + 4 )}

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

Expand telescoping series:

\frac{25}{48}

= \frac{25}{48}

n = 0 \sum \infty \frac{1}{100 + n}

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

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

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

n = 1 \sum \infty \frac{9}{n ^{2} + 25}

What is the sum from n=1 to infinity of 1/(n^2+4n) ?
The sum from n=1 to infinity of 1/(n^2+4n) is 25/48