What is the sum from n=0 to infinity of 2/(7-n) ?

The sum from n=0 to infinity of 2/(7-n) is diverges

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sum from n=0 to infinity of 2/(7-n)

n = 0 \sum \infty \frac{2}{7 - n}

diverges

Solution steps

n = 8 \sum \infty \frac{2}{7 - n}

Apply the constant multiplication rule:

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

= 2 \cdot n = 8 \sum \infty \frac{1}{7 - n}

Factor

7 - n : - (- 7 + n)

= 2 \cdot n = 8 \sum \infty \frac{1}{- ( - 7 + n )}

Apply the constant multiplication rule:

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

= 2 (- n = 8 \sum \infty \frac{1}{- 7 + n})

Apply Series Comparison Test:

diverges

= 2 (- diverges)

= diverges

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

n = 1 \sum \infty \frac{2}{3 ^{n} + 20}

n = 1 \sum \infty \frac{7}{n ^{2} + n}

i = 1 \sum \infty 5 \cdot i

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

What is the sum from n=0 to infinity of 2/(7-n) ?
The sum from n=0 to infinity of 2/(7-n) is diverges