What is the sum from n=1 to infinity of 7/(6n) ?

The sum from n=1 to infinity of 7/(6n) is diverges

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sum from n=1 to infinity of 7/(6n)

n = 1 \sum \infty \frac{7}{6 n}

diverges

Solution steps

n = 1 \sum \infty \frac{7}{6 n}

Apply the constant multiplication rule:

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

= \frac{7}{6} \cdot n = 1 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= \frac{7}{6} diverges

= diverges

k = 1 \sum \infty tan (\frac{1}{k})

n = 1 \sum \infty \frac{1}{25 n ^{2} - 1}

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

n = 1 \sum \infty \frac{sin ( 1 )}{n}

n = 1 \sum \infty \frac{n}{6 n ^{3} + 5}

What is the sum from n=1 to infinity of 7/(6n) ?
The sum from n=1 to infinity of 7/(6n) is diverges