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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

f (x) = arcsec (5 x)

\int x os (2 x) d x

\frac{d}{d x} (a sin (b x + c) + d)

\int 5 e^{a x} d x

\int_{0}^{6} (6 - x) d x

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