What is the sum from n=0 to infinity of 1/(1-4n) ?

The sum from n=0 to infinity of 1/(1-4n) is diverges

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sum from n=0 to infinity of 1/(1-4n)

n = 0 \sum \infty \frac{1}{1 - 4 n}

diverges

Solution steps

n = 0 \sum \infty \frac{1}{1 - 4 n}

Factor

1 - 4 n : - (- 1 + 4 n)

= n = 0 \sum \infty \frac{1}{- ( - 1 + 4 n )}

Apply the constant multiplication rule:

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

= - n = 0 \sum \infty \frac{1}{- 1 + 4 n}

Apply Series Integral Test:

diverges

= - diverges

= diverges

n = 1 \sum \infty \frac{9}{n ^{2} + 3 n}

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

n = 0 \sum \infty \frac{1}{9 - n}

n = 0 \sum \infty \frac{1}{1 - 7 n}

n = 0 \sum \infty \frac{1}{8 n + 4}

What is the sum from n=0 to infinity of 1/(1-4n) ?
The sum from n=0 to infinity of 1/(1-4n) is diverges