What is the sum from n=0 to infinity of 4/(7-n) ?

The sum from n=0 to infinity of 4/(7-n) is diverges

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sum from n=0 to infinity of 4/(7-n)

n = 0 \sum \infty \frac{4}{7 - n}

diverges

Solution steps

n = 8 \sum \infty \frac{4}{7 - n}

Apply the constant multiplication rule:

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

= 4 \cdot n = 8 \sum \infty \frac{1}{7 - n}

Factor

7 - n : - (- 7 + n)

= 4 \cdot n = 8 \sum \infty \frac{1}{- ( - 7 + n )}

Apply the constant multiplication rule:

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

= 4 (- n = 8 \sum \infty \frac{1}{- 7 + n})

Apply Series Comparison Test:

diverges

= 4 (- diverges)

= diverges

n = 1 \sum \infty \frac{7 ^{n}}{n + 1}

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

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

n = 0 \sum \infty \frac{1}{5} (\frac{4}{5})^{n}

n = 1 \sum \infty \frac{1}{n ^{2.5}}

What is the sum from n=0 to infinity of 4/(7-n) ?
The sum from n=0 to infinity of 4/(7-n) is diverges