What is the sum from n=0 to infinity of 1/(9-n) ?

The sum from n=0 to infinity of 1/(9-n) is diverges

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sum from n=0 to infinity of 1/(9-n)

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

diverges

Solution steps

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

Factor

9 - n : - (- 9 + n)

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

diverges

= - diverges

= diverges

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

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

n = 0 \sum \infty \frac{n !}{9 ^{n}}

n = 1 \sum \infty \frac{9}{n ^{2} + 16}

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

What is the sum from n=0 to infinity of 1/(9-n) ?
The sum from n=0 to infinity of 1/(9-n) is diverges