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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

converges

= 7 converges

= converges

n = 1 \sum \infty \frac{8 ^{n}}{n}

n = 2 \sum \infty (\frac{5}{6})^{2 n}

k = 5 \sum \infty \frac{1}{k ^{2} - k}

n = 0 \sum \infty (2)^{n}

n = 1 \sum \infty \frac{3 ^{n}}{n + 1}

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