What is the sum from n=0 to infinity of 1/(6-n) ?

The sum from n=0 to infinity of 1/(6-n) is diverges

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sum from n=0 to infinity of 1/(6-n)

n = 0 \sum \infty \frac{1}{6 - n}

diverges

Solution steps

n = 7 \sum \infty \frac{1}{6 - n}

Factor

6 - n : - (- 6 + n)

= n = 7 \sum \infty \frac{1}{- ( - 6 + n )}

Apply the constant multiplication rule:

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

= - n = 7 \sum \infty \frac{1}{- 6 + n}

Apply Series Comparison Test:

diverges

= - diverges

= diverges

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

n = 1 \sum \infty 6 (0.8)^{n - 1}

n = 1 \sum \infty (cos (21))^{n}

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

What is the sum from n=0 to infinity of 1/(6-n) ?
The sum from n=0 to infinity of 1/(6-n) is diverges