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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Divergence Test:

diverges

= - diverges

= diverges

n = 0 \sum \infty 9 e^{- 9 n}

n = 0 \sum \infty 10 (\frac{3}{4})^{n}

n = 0 \sum \infty \frac{1}{2 ^{3 n}}

n = 0 \sum \infty \frac{5}{1 0 ^{n}}

k = 1 \sum \infty \frac{k}{ln ( k )}

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