What is the sum from n=0 to infinity of 2/(3-n) ?

The sum from n=0 to infinity of 2/(3-n) is diverges

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sum from n=0 to infinity of 2/(3-n)

n = 0 \sum \infty \frac{2}{3 - n}

diverges

Solution steps

n = 4 \sum \infty \frac{2}{3 - n}

Apply the constant multiplication rule:

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

= 2 \cdot n = 4 \sum \infty \frac{1}{3 - n}

Factor

3 - n : - (- 3 + n)

= 2 \cdot n = 4 \sum \infty \frac{1}{- ( - 3 + n )}

Apply the constant multiplication rule:

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

= 2 (- n = 4 \sum \infty \frac{1}{- 3 + n})

Apply Series Comparison Test:

diverges

= 2 (- diverges)

= diverges

n = 1 \sum \infty - 3 (\frac{8}{7})^{n - 1}

n = 0 \sum \infty \frac{1 + ( - 1 ) ^{n}}{n}

n = 0 \sum \infty 8^{\frac{1}{n}}

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

n = 1 \sum \infty \frac{1}{( 7 n + 4 ) ^{2}}

What is the sum from n=0 to infinity of 2/(3-n) ?
The sum from n=0 to infinity of 2/(3-n) is diverges