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

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

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

n = 1 \sum \infty \frac{n - 9}{n + 9}

diverges

Solution steps

n = 1 \sum \infty \frac{n - 9}{n + 9}

Apply Series Divergence Test:

diverges

= diverges

\frac{( 11 s ^{2} - 2 s + 5 )}{( s - 2 ) ( 2 s - 1 ) ( s + 1 )}

- 3 t

\frac{d}{d x} (\frac{2 x ^{5} + 3 x}{x ^{4}})

\int (5 x^{4}) d x

\int 3 x + \frac{1}{x} d x

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