What is the sum from n=1 to infinity of 4/(3n) ?

The sum from n=1 to infinity of 4/(3n) is diverges

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sum from n=1 to infinity of 4/(3n)

n = 1 \sum \infty \frac{4}{3 n}

diverges

Solution steps

n = 1 \sum \infty \frac{4}{3 n}

Apply the constant multiplication rule:

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

= \frac{4}{3} \cdot n = 1 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= \frac{4}{3} diverges

= diverges

x \to 0 lim (x^{4} cos (\frac{9}{x}))

t - sin (t)

ln (2 x + 1)

\int_{0}^{\frac{3}{4}} 30 x d x

f (x) = x + 9 x

What is the sum from n=1 to infinity of 4/(3n) ?
The sum from n=1 to infinity of 4/(3n) is diverges