What is the sum from n=1 to infinity of (2n)/(n+3) ?

The sum from n=1 to infinity of (2n)/(n+3) is diverges

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sum from n=1 to infinity of (2n)/(n+3)

n = 1 \sum \infty \frac{2 n}{n + 3}

diverges

Solution steps

n = 1 \sum \infty \frac{2 n}{n + 3}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{n}{n + 3}

Apply Series Divergence Test:

diverges

= 2 diverges

= diverges

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

y = x^{2}, x + y = 2

y^{^{'}} + \frac{1}{2} \cdot y = \frac{3}{2} \cdot t

B (x - 2)^{2} x^{2}

\int \frac{2 x}{x + 5} d x

What is the sum from n=1 to infinity of (2n)/(n+3) ?
The sum from n=1 to infinity of (2n)/(n+3) is diverges