What is the sum from n=1 to infinity of n/((3+n)) ?

The sum from n=1 to infinity of n/((3+n)) is diverges

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sum from n=1 to infinity of n/((3+n))

n = 1 \sum \infty \frac{n}{( 3 + n )}

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

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\frac{d}{d x} (\frac{1 + x ^{2}}{1 + x})

\frac{16}{( 1 - x ) ^{3}}

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What is the sum from n=1 to infinity of n/((3+n)) ?
The sum from n=1 to infinity of n/((3+n)) is diverges