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

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

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

n = 1 \sum \infty \frac{ln ( n )}{5 n}

diverges

Solution steps

n = 1 \sum \infty \frac{ln ( n )}{5 n}

Apply the constant multiplication rule:

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

= \frac{1}{5} \cdot n = 1 \sum \infty \frac{ln ( n )}{n}

Apply Series Integral Test:

diverges

= \frac{1}{5} diverges

= diverges

x \to 1 lim (- cos (\frac{1}{x - 1}))

\frac{\partial}{\partial x} (6 e^{x} cos (yz))

\frac{d}{d x} (5 + sin (45))

x \to 0 + lim (\frac{x ^{2}}{5} - \frac{7}{x})

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