What is the sum from n=0 to infinity of 4/(5-n) ?

The sum from n=0 to infinity of 4/(5-n) is diverges

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sum from n=0 to infinity of 4/(5-n)

n = 0 \sum \infty \frac{4}{5 - n}

diverges

Solution steps

n = 6 \sum \infty \frac{4}{5 - n}

Apply the constant multiplication rule:

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

= 4 \cdot n = 6 \sum \infty \frac{1}{5 - n}

Factor

5 - n : - (- 5 + n)

= 4 \cdot n = 6 \sum \infty \frac{1}{- ( - 5 + n )}

Apply the constant multiplication rule:

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

= 4 (- n = 6 \sum \infty \frac{1}{- 5 + n})

Apply Series Comparison Test:

diverges

= 4 (- diverges)

= diverges

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

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

n = 0 \sum \infty (1 + \frac{2}{5 n})^{2}

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

n = 1 \sum \infty \frac{n}{n + 200}

What is the sum from n=0 to infinity of 4/(5-n) ?
The sum from n=0 to infinity of 4/(5-n) is diverges