What is the sum from n=0 to infinity of 4n^2 ?

The sum from n=0 to infinity of 4n^2 is diverges

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sum from n=0 to infinity of 4n^2

n = 0 \sum \infty 4 n^{2}

diverges

Solution steps

n = 0 \sum \infty 4 n^{2}

Apply the constant multiplication rule:

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

= 4 \cdot n = 0 \sum \infty n^{2}

Apply Series Divergence Test:

diverges

= 4 diverges

= diverges

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x \to 0 lim (\frac{( 1 - x ) ^{n} - 1}{x})

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What is the sum from n=0 to infinity of 4n^2 ?
The sum from n=0 to infinity of 4n^2 is diverges