What is the sum from n=0 to infinity of 1/(1-5n) ?

The sum from n=0 to infinity of 1/(1-5n) is diverges

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sum from n=0 to infinity of 1/(1-5n)

n = 0 \sum \infty \frac{1}{1 - 5 n}

diverges

Solution steps

n = 0 \sum \infty \frac{1}{1 - 5 n}

Factor

1 - 5 n : - (- 1 + 5 n)

= n = 0 \sum \infty \frac{1}{- ( - 1 + 5 n )}

Apply the constant multiplication rule:

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

= - n = 0 \sum \infty \frac{1}{- 1 + 5 n}

Apply Series Integral Test:

diverges

= - diverges

= diverges

n = 1 \sum \infty (\frac{3}{7})^{n - 1}

n = 6 \sum \infty \frac{1}{n} - \frac{1}{n + 2}

n = 0 \sum \infty c n

n = 1 \sum \infty - 5 n

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

What is the sum from n=0 to infinity of 1/(1-5n) ?
The sum from n=0 to infinity of 1/(1-5n) is diverges