What is the sum from k=1 to infinity of 4/k ?

The sum from k=1 to infinity of 4/k is diverges

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sum from k=1 to infinity of 4/k

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

= 4 \cdot k = 1 \sum \infty \frac{1}{k}

Apply Cauchy's Convergence Condition:

diverges

= 4 diverges

= diverges

n = 1 \sum \infty n (- \frac{3}{7})^{n}

n = 1 \sum \infty \frac{5}{n ^{2} + 36}

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

k = 1 \sum \infty (cos (10))^{k}

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

What is the sum from k=1 to infinity of 4/k ?
The sum from k=1 to infinity of 4/k is diverges