What is the sum from n=1 to infinity of 2/(3n+3) ?

The sum from n=1 to infinity of 2/(3n+3) is diverges

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sum from n=1 to infinity of 2/(3n+3)

n = 1 \sum \infty \frac{2}{3 n + 3}

diverges

Solution steps

n = 1 \sum \infty \frac{2}{3 n + 3}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{1}{3 n + 3}

Apply Series Integral Test:

diverges

= 2 diverges

= diverges

n = 1 \sum \infty \frac{6}{n} - \frac{6}{n ^{2}}

n = 1 \sum \infty 30 (0.1)^{n}

n = 1 \sum \infty \frac{8}{n} + 3^{- n}

n = 1 \sum \infty \frac{1}{( 9 n ) ^{9}}

k = 2 \sum \infty (- 0.15)^{k}

What is the sum from n=1 to infinity of 2/(3n+3) ?
The sum from n=1 to infinity of 2/(3n+3) is diverges