What is the sum from n=0 to infinity of 2/(7-3n) ?

The sum from n=0 to infinity of 2/(7-3n) is diverges

Popular >

sum from n=0 to infinity of 2/(7-3n)

n = 0 \sum \infty \frac{2}{7 - 3 n}

diverges

Solution steps

n = 0 \sum \infty \frac{2}{7 - 3 n}

Apply the constant multiplication rule:

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

= 2 \cdot n = 0 \sum \infty \frac{1}{7 - 3 n}

Factor

7 - 3 n : - (- 7 + 3 n)

= 2 \cdot n = 0 \sum \infty \frac{1}{- ( - 7 + 3 n )}

Apply the constant multiplication rule:

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

= 2 (- n = 0 \sum \infty \frac{1}{- 7 + 3 n})

Apply Series Integral Test:

diverges

= 2 (- diverges)

= diverges

n = 0 \sum \infty \frac{n}{n}

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

n = 0 \sum \infty 5 \cdot (\frac{2}{3})^{n}

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

n = 600 \sum \infty \frac{1}{n}

What is the sum from n=0 to infinity of 2/(7-3n) ?
The sum from n=0 to infinity of 2/(7-3n) is diverges