What is the sum from n=0 to infinity of (3n+1)/(2n) ?

The sum from n=0 to infinity of (3n+1)/(2n) is diverges

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sum from n=0 to infinity of (3n+1)/(2n)

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Divergence Test:

diverges

= \frac{1}{2} diverges

= diverges

n = 1 \sum \infty \frac{1}{ln ( 4 ^{n} )}

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

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

n = 0 \sum \infty (\frac{11}{3})^{n}

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

What is the sum from n=0 to infinity of (3n+1)/(2n) ?
The sum from n=0 to infinity of (3n+1)/(2n) is diverges