What is the sum from n=0 to infinity of 3/(5-n) ?

The sum from n=0 to infinity of 3/(5-n) is diverges

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sum from n=0 to infinity of 3/(5-n)

n = 0 \sum \infty \frac{3}{5 - n}

diverges

Solution steps

n = 6 \sum \infty \frac{3}{5 - n}

Apply the constant multiplication rule:

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

= 3 \cdot n = 6 \sum \infty \frac{1}{5 - n}

Factor

5 - n : - (- 5 + n)

= 3 \cdot n = 6 \sum \infty \frac{1}{- ( - 5 + n )}

Apply the constant multiplication rule:

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

= 3 (- n = 6 \sum \infty \frac{1}{- 5 + n})

Apply Series Comparison Test:

diverges

= 3 (- diverges)

= diverges

n = 2 \sum \infty \frac{4}{n ln ( n )}

n = 1 \sum \infty \frac{6 ^{n}}{n}

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

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

n = 0 \sum \infty ln (5 - n)

What is the sum from n=0 to infinity of 3/(5-n) ?
The sum from n=0 to infinity of 3/(5-n) is diverges