What is the sum from n=0 to infinity of 1/(3-n) ?

The sum from n=0 to infinity of 1/(3-n) is diverges

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sum from n=0 to infinity of 1/(3-n)

n = 0 \sum \infty \frac{1}{3 - n}

diverges

Solution steps

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

Factor

3 - n : - (- 3 + n)

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

diverges

= - diverges

= diverges

n = 1 \sum \infty \frac{3}{2 n}

k = 1 \sum \infty k e^{- 7 k}

n = 0 \sum \infty 2 (\frac{3}{2})^{n}

k = 0 \sum \infty (\frac{2}{3})^{k}

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

What is the sum from n=0 to infinity of 1/(3-n) ?
The sum from n=0 to infinity of 1/(3-n) is diverges