What is the sum from k=1 to infinity of 3/k ?

The sum from k=1 to infinity of 3/k is diverges

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sum from k=1 to infinity of 3/k

k = 1 \sum \infty \frac{3}{k}

diverges

Solution steps

k = 1 \sum \infty \frac{3}{k}

Apply the constant multiplication rule:

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

= 3 \cdot k = 1 \sum \infty \frac{1}{k}

Apply Cauchy's Convergence Condition:

diverges

= 3 diverges

= diverges

n = 3 \sum \infty \frac{1}{n ^{2} ln ( n )}

n = 2 \sum \infty \frac{5}{n ln ( n )}

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

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

n = 2 \sum \infty (- 0.6)^{n}

What is the sum from k=1 to infinity of 3/k ?
The sum from k=1 to infinity of 3/k is diverges