What is the sum from n=0 to infinity of 8/(5-n) ?

The sum from n=0 to infinity of 8/(5-n) is diverges

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sum from n=0 to infinity of 8/(5-n)

n = 0 \sum \infty \frac{8}{5 - n}

diverges

Solution steps

n = 6 \sum \infty \frac{8}{5 - n}

Apply the constant multiplication rule:

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

= 8 \cdot n = 6 \sum \infty \frac{1}{5 - n}

Factor

5 - n : - (- 5 + n)

= 8 \cdot n = 6 \sum \infty \frac{1}{- ( - 5 + n )}

Apply the constant multiplication rule:

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

= 8 (- n = 6 \sum \infty \frac{1}{- 5 + n})

Apply Series Comparison Test:

diverges

= 8 (- diverges)

= diverges

n = 1 \sum \infty (cos (18))^{n}

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

n = 0 \sum \infty 8^{- n}

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

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

What is the sum from n=0 to infinity of 8/(5-n) ?
The sum from n=0 to infinity of 8/(5-n) is diverges