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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= \frac{1}{2} diverges

= diverges

n = 1 \sum \infty (\frac{6 π}{7})^{n}

n = 4 \sum \infty \frac{1}{n ^{2} - n}

n = 0 \sum \infty (- 9)^{n}

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

n = 0 \sum \infty 8 e^{- n}

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