What is the sum from n=1 to infinity of n^3+n ?

The sum from n=1 to infinity of n^3+n is diverges

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sum from n=1 to infinity of n^3+n

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

(x^{5})^{^{'}}

\int \frac{1}{4 x + 9} d x

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\int_{0}^{π} (π - x) sin (n x) d x

\frac{\partial}{\partial x} (x y^{2} - x^{3} y)

What is the sum from n=1 to infinity of n^3+n ?
The sum from n=1 to infinity of n^3+n is diverges