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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Divergence Test:

diverges

= \frac{2}{3} diverges

= diverges

\frac{\partial}{\partial x} (3 e^{y})

\frac{\partial}{\partial x} (sin (2 x) cos (2 t))

\int \frac{ln ( 2 x )}{ln ( 4 x ) x} d x

\frac{d}{d x} ((x + 6)^{7} (x + 3)^{5})

y^{^{'}} = \frac{( x - 3 )}{( y - 4 )}

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