What is the sum from n=1 to infinity of 1/(9n) ?

The sum from n=1 to infinity of 1/(9n) is diverges

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sum from n=1 to infinity of 1/(9n)

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

= \frac{1}{9} \cdot n = 1 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= \frac{1}{9} diverges

= diverges

k = 2 \sum \infty (\frac{3}{5})^{3 k}

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

k = 1 \sum \infty \frac{1}{2 k}

n = 0 \sum \infty \frac{1}{4} (\frac{3}{4})^{n}

n = 1 \sum \infty 12 sin (\frac{1}{n})

What is the sum from n=1 to infinity of 1/(9n) ?
The sum from n=1 to infinity of 1/(9n) is diverges