What is the sum from n=0 to infinity of 4/(3n+2) ?

The sum from n=0 to infinity of 4/(3n+2) is diverges

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sum from n=0 to infinity of 4/(3n+2)

n = 0 \sum \infty \frac{4}{3 n + 2}

diverges

Solution steps

n = 0 \sum \infty \frac{4}{3 n + 2}

Apply the constant multiplication rule:

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

= 4 \cdot n = 0 \sum \infty \frac{1}{3 n + 2}

Apply Series Integral Test:

diverges

= 4 diverges

= diverges

n = 1 \sum \infty \frac{1}{12 n + 7}

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

n = 2 \sum \infty (\frac{5}{7})^{2 n}

n = 0 \sum \infty \frac{1}{5 ^{n} + 3}

n = 0 \sum \infty \frac{n + 1}{1 0 ^{n}}

What is the sum from n=0 to infinity of 4/(3n+2) ?
The sum from n=0 to infinity of 4/(3n+2) is diverges