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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Simplify

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

Every infinite sum of a non-zero constant diverges

= \frac{2}{3} diverges

= diverges

1 + x^{2}

\int_{0}^{9} 4 x - \frac{4 x}{3} d x

\frac{\partial}{\partial x} (3 x^{3} y^{2} z)

0.03 (1 - cos (2 t))

\int_{0}^{\infty} 3 e^{2 x} d x

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