What is the sum from k=1 to infinity of 6/k ?

The sum from k=1 to infinity of 6/k is diverges

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sum from k=1 to infinity of 6/k

k = 1 \sum \infty \frac{6}{k}

diverges

Solution steps

k = 1 \sum \infty \frac{6}{k}

Apply the constant multiplication rule:

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

= 6 \cdot k = 1 \sum \infty \frac{1}{k}

Apply Cauchy's Convergence Condition:

diverges

= 6 diverges

= diverges

n = 1 \sum \infty \frac{1}{3 n ^{2} + 9}

n = 1 \sum \infty \frac{1}{3 n ^{2} + 4}

n = 1 \sum \infty \frac{7}{n ^{9}}

n = 0 \sum \infty 1. 1^{n}

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

What is the sum from k=1 to infinity of 6/k ?
The sum from k=1 to infinity of 6/k is diverges