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

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

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

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

diverges

Solution steps

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

Apply Cauchy's Convergence Condition:

diverges

= \frac{1}{7} diverges

= diverges

n = 1 \sum \infty (\frac{5}{4})^{n - 1}

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

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

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

n = 1 \sum \infty π

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