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

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

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

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

diverges

Solution steps

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

Simplify

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

= n = 1 \sum \infty n

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty \frac{26}{n !}

k = 1 \sum \infty (- \frac{1}{2})^{k}

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

n = 1 \sum \infty 1 - e^{- n}

n = 1 \sum \infty \frac{1 7 ^{n}}{n !}

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