What is the sum from n=1 to infinity of 8/n ?

The sum from n=1 to infinity of 8/n is diverges

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sum from n=1 to infinity of 8/n

n = 1 \sum \infty \frac{8}{n}

diverges

Solution steps

n = 1 \sum \infty \frac{8}{n}

Apply the constant multiplication rule:

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

= 8 \cdot n = 1 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= 8 diverges

= diverges

n = 1 \sum \infty \frac{1}{n ^{\frac{1}{5}}}

n = 0 \sum \infty \frac{n !}{7 ^{n}}

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

n = 1 \sum \infty n e^{- 9 n}

n = 1 \sum \infty 5^{1 - n}

What is the sum from n=1 to infinity of 8/n ?
The sum from n=1 to infinity of 8/n is diverges