What is the sum from n=0 to infinity of 8/(9-n) ?

The sum from n=0 to infinity of 8/(9-n) is diverges

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sum from n=0 to infinity of 8/(9-n)

n = 0 \sum \infty \frac{8}{9 - n}

diverges

Solution steps

n = 10 \sum \infty \frac{8}{9 - n}

Apply the constant multiplication rule:

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

= 8 \cdot n = 10 \sum \infty \frac{1}{9 - n}

Factor

9 - n : - (- 9 + n)

= 8 \cdot n = 10 \sum \infty \frac{1}{- ( - 9 + n )}

Apply the constant multiplication rule:

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

= 8 (- n = 10 \sum \infty \frac{1}{- 9 + n})

Apply Series Comparison Test:

diverges

= 8 (- diverges)

= diverges

n = 0 \sum \infty \frac{- 1 ^{n}}{3 ^{n}}

n = 0 \sum \infty n^{- 9}

n = 1 \sum \infty (\frac{4}{9})^{n}

n = 0 \sum \infty n^{- n}

n = 1 \sum \infty (\frac{π}{e})^{n}

What is the sum from n=0 to infinity of 8/(9-n) ?
The sum from n=0 to infinity of 8/(9-n) is diverges