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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Divergence Test:

diverges

= 3 diverges

= diverges

\frac{\partial}{\partial x} (\frac{y}{x ^{2} + y ^{2} + 1})

\frac{\partial}{\partial v} (u^{2})

\frac{\partial}{\partial x} (y + x^{2})

x y^{^{'}} + 2 y = - 4 x

\frac{d}{d x} ((x^{3})^{5})

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