What is the sum from n=0 to infinity of 3/(3-6n) ?

The sum from n=0 to infinity of 3/(3-6n) is diverges

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sum from n=0 to infinity of 3/(3-6n)

n = 0 \sum \infty \frac{3}{3 - 6 n}

diverges

Solution steps

n = 0 \sum \infty \frac{3}{3 - 6 n}

Apply the constant multiplication rule:

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

= 3 \cdot n = 0 \sum \infty \frac{1}{3 - 6 n}

Factor

3 - 6 n : - (- 3 + 6 n)

= 3 \cdot n = 0 \sum \infty \frac{1}{- ( - 3 + 6 n )}

Apply the constant multiplication rule:

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

= 3 (- n = 0 \sum \infty \frac{1}{- 3 + 6 n})

Apply Series Integral Test:

diverges

= 3 (- diverges)

= diverges

What is the sum from n=0 to infinity of 3/(3-6n) ?
The sum from n=0 to infinity of 3/(3-6n) is diverges