What is the sum from n=1 to infinity of (ln(n))/(7n) ?

The sum from n=1 to infinity of (ln(n))/(7n) is diverges

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sum from n=1 to infinity of (ln(n))/(7n)

n = 1 \sum \infty \frac{ln ( n )}{7 n}

diverges

Solution steps

n = 1 \sum \infty \frac{ln ( n )}{7 n}

Apply the constant multiplication rule:

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

= \frac{1}{7} \cdot n = 1 \sum \infty \frac{ln ( n )}{n}

Apply Series Integral Test:

diverges

= \frac{1}{7} diverges

= diverges

n = 4 \sum \infty \frac{5}{3 ^{n}}

n = 0 \sum \infty 7 - n^{4}

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

n = 1 \sum \infty \frac{3}{n ^{10}}

k = 2 \sum \infty (- 0.1)^{k}

What is the sum from n=1 to infinity of (ln(n))/(7n) ?
The sum from n=1 to infinity of (ln(n))/(7n) is diverges