What is the sum from n=1 to infinity of 4/(n(n+7)) ?

The sum from n=1 to infinity of 4/(n(n+7)) is converges

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sum from n=1 to infinity of 4/(n(n+7))

n = 1 \sum \infty \frac{4}{n ( n + 7 )}

converges

Solution steps

n = 1 \sum \infty \frac{4}{n ( n + 7 )}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty \frac{1}{n ( n + 7 )}

Apply Series Comparison Test:

converges

= 4 converges

= converges

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

f (x) = \frac{x ^{2} + 3 x - 2}{x ^{2} - 5 x + 6}

f (x) = e^{x} - x^{3}

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (1) = 0, y^{^{'}} (1) = - 1

y^{^{'}} = \frac{( x - y )}{( x + 2 y )}, y (0) = 1

What is the sum from n=1 to infinity of 4/(n(n+7)) ?
The sum from n=1 to infinity of 4/(n(n+7)) is converges