What is the sum from n=0 to infinity of 4/(3-n) ?

The sum from n=0 to infinity of 4/(3-n) is diverges

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sum from n=0 to infinity of 4/(3-n)

n = 0 \sum \infty \frac{4}{3 - n}

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Factor

3 - n : - (- 3 + n)

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

diverges

= 4 (- diverges)

= diverges

n = 0 \sum \infty \frac{1}{n ^{4} + n ^{2}}

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

k = 2 \sum \infty (- 0.35)^{k}

n = 1 \sum \infty \frac{7 ^{n}}{1 0 ^{n}}

k = 5 \sum \infty 2 (- \frac{1}{4})^{k}

What is the sum from n=0 to infinity of 4/(3-n) ?
The sum from n=0 to infinity of 4/(3-n) is diverges