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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Cauchy's Convergence Condition:

diverges

= \frac{1}{6} diverges

= diverges

y = x^{2} + 5 x, x = 8

\int (x^{2} - 2 x + 3) d x

\int \frac{3 x ^{3}}{x ^{2} + 1} d x

n = 0 \sum \infty \frac{1}{n}

\frac{\partial}{\partial x} (x^{0.7} y^{0.3})

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