What is the sum from n=1 to infinity of n/(5n) ?

The sum from n=1 to infinity of n/(5n) is diverges

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sum from n=1 to infinity of n/(5n)

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Simplify

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

Every infinite sum of a non-zero constant diverges

= \frac{1}{5} diverges

= diverges

n = 0 \sum \infty (\frac{2 - 1}{2})^{n}

n = 1 \sum \infty (- 1)^{n} 0

n = 4 \sum \infty \frac{4}{( n ^{2} - 1 )}

n = 1 \sum \infty \frac{n + 8}{n ^{2}}

n = 0 \sum \infty \frac{i ^{2 n}}{4 n}

What is the sum from n=1 to infinity of n/(5n) ?
The sum from n=1 to infinity of n/(5n) is diverges