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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Divergence Test:

diverges

= \frac{1}{3} diverges

= diverges

(3) (6.9)

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