What is the sum from n=0 to infinity of (n^2)/(5n) ?

The sum from n=0 to infinity of (n^2)/(5n) is diverges

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sum from n=0 to infinity of (n^2)/(5n)

n = 0 \sum \infty \frac{n ^{2}}{5 n}

diverges

Solution steps

n = 1 \sum \infty \frac{n ^{2}}{5 n}

Apply the constant multiplication rule:

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

= \frac{1}{5} \cdot n = 1 \sum \infty \frac{n ^{2}}{n}

Apply Series Divergence Test:

diverges

= \frac{1}{5} diverges

= diverges

6 x^{4}

\frac{\partial}{\partial u} (w arctan (uv))

\int sinh (t) d t

\int \frac{x ^{4}}{1 - x} d x

\frac{\partial}{\partial x} (y - z)

What is the sum from n=0 to infinity of (n^2)/(5n) ?
The sum from n=0 to infinity of (n^2)/(5n) is diverges