What is the sum from n=0 to infinity of 8/((1-7n)) ?

The sum from n=0 to infinity of 8/((1-7n)) is diverges

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sum from n=0 to infinity of 8/((1-7n))

n = 0 \sum \infty \frac{8}{( 1 - 7 n )}

diverges

Solution steps

n = 0 \sum \infty \frac{8}{1 - 7 n}

Apply the constant multiplication rule:

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

= 8 \cdot n = 0 \sum \infty \frac{1}{1 - 7 n}

Factor

1 - 7 n : - (- 1 + 7 n)

= 8 \cdot n = 0 \sum \infty \frac{1}{- ( - 1 + 7 n )}

Apply the constant multiplication rule:

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

= 8 (- n = 0 \sum \infty \frac{1}{- 1 + 7 n})

Apply Series Integral Test:

diverges

= 8 (- diverges)

= diverges

n = 1 \sum \infty 53 (\frac{1}{2})^{n}

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

n = 0 \sum \infty \frac{1}{( n ln ( n ) )}

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

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

What is the sum from n=0 to infinity of 8/((1-7n)) ?
The sum from n=0 to infinity of 8/((1-7n)) is diverges