What is the sum from n=0 to infinity of 1/(4-n) ?

The sum from n=0 to infinity of 1/(4-n) is diverges

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sum from n=0 to infinity of 1/(4-n)

n = 0 \sum \infty \frac{1}{4 - n}

diverges

Solution steps

n = 5 \sum \infty \frac{1}{4 - n}

Factor

4 - n : - (- 4 + n)

= n = 5 \sum \infty \frac{1}{- ( - 4 + n )}

Apply the constant multiplication rule:

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

= - n = 5 \sum \infty \frac{1}{- 4 + n}

Apply Series Comparison Test:

diverges

= - diverges

= diverges

n = 1 \sum \infty \frac{9}{n ( n + 2 )}

n = 1 \sum \infty \frac{4}{( - 3 ) ^{n}}

n = 0 \sum \infty \frac{2}{2 ^{n}}

n = 0 \sum \infty ln (1 - n)

k = 0 \sum \infty \frac{k}{k + 1}

What is the sum from n=0 to infinity of 1/(4-n) ?
The sum from n=0 to infinity of 1/(4-n) is diverges