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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

f (x) = (2 + 4 x) (5 - 4 x), at x = 1

y = 2 x^{3} - 2 x^{2} - 2 x

\frac{d y}{d x} + \frac{y}{x} = x^{3}

\frac{d}{d x} (\frac{e ^{a x^{2} + b x + c}}{x})

e \to 1 lim (5 ln^{2} (e))

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