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

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

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

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

diverges

Solution steps

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

Factor

8 - n : - (- 8 + n)

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

diverges

= - diverges

= diverges

n = 1 \sum \infty 9 (0.1)^{n - 1}

n = 2 \sum \infty \frac{n ^{2}}{3 ^{n}}

n = 1 \sum \infty 3 (0.3)^{n - 1}

n = 1 \sum \infty \frac{5 n ^{3}}{6 ^{n}}

k = 0 \sum \infty (\frac{1}{3})^{k}

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