What is the sum from n=1 to infinity of 9/(n(n+9)) ?

The sum from n=1 to infinity of 9/(n(n+9)) is converges

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sum from n=1 to infinity of 9/(n(n+9))

n = 1 \sum \infty \frac{9}{n ( n + 9 )}

converges

Solution steps

n = 1 \sum \infty \frac{9}{n ( n + 9 )}

Apply the constant multiplication rule:

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

= 9 \cdot n = 1 \sum \infty \frac{1}{n ( n + 9 )}

Apply Series Comparison Test:

converges

= 9 converges

= converges

n = 1 \sum \infty e^{an}

n = 1 \sum \infty n^{4} e^{- n}

n = 1 \sum \infty \frac{1}{n ^{\frac{7}{6}}}

n = 0 \sum \infty 10 + 2 n

n = 1 \sum \infty n (\frac{6}{5})^{n}

What is the sum from n=1 to infinity of 9/(n(n+9)) ?
The sum from n=1 to infinity of 9/(n(n+9)) is converges