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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Simplify

\frac{n ^{3}}{n ^{2}} : n

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

Apply Series Divergence Test:

diverges

= \frac{1}{3} diverges

= diverges

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