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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

x \to 1 lim ((x^{2} - 1)^{x^{2}})

y^{^{'}} = \frac{x y}{2}

\int_{0}^{1} x^{4} e^{5 x} d x

\int a tan^{2} (θ) d θ

\frac{d}{d x} (4 x + ln (x + 1))

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