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

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

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

n = 1 \sum \infty \frac{ln ( n )}{3 n}

diverges

Solution steps

n = 1 \sum \infty \frac{ln ( n )}{3 n}

Apply the constant multiplication rule:

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

= \frac{1}{3} \cdot n = 1 \sum \infty \frac{ln ( n )}{n}

Apply Series Integral Test:

diverges

= \frac{1}{3} diverges

= diverges

n = 0 \sum \infty \frac{n}{ln ( n ^{2} )}

n = 1 \sum \infty n^{- s i n (1)}

n = 1 \sum \infty cos^{n} (5)

k = 1 \sum \infty (\frac{2}{5})^{k}

n = 5 \sum \infty \frac{1}{ln ( n - 3 )}

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