What is the sum from n=1 to infinity of 4/(5n) ?

The sum from n=1 to infinity of 4/(5n) is diverges

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sum from n=1 to infinity of 4/(5n)

n = 1 \sum \infty \frac{4}{5 n}

diverges

Solution steps

n = 1 \sum \infty \frac{4}{5 n}

Apply the constant multiplication rule:

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

= \frac{4}{5} \cdot n = 1 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= \frac{4}{5} diverges

= diverges

\frac{\partial}{\partial s} (\frac{( 3 r + s )}{t})

(- 7.2) (8.2)

\int_{0}^{8 π} x^{2} + 5 d x

x^{2} \frac{d y}{d x} + x (x + 2) y = e^{x}

\int \frac{5}{( x ^{2} + 4 )} d x

What is the sum from n=1 to infinity of 4/(5n) ?
The sum from n=1 to infinity of 4/(5n) is diverges