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

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

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

n = 1 \sum \infty (n)^{2}

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty n tan (\frac{1}{n})

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

n = 0 \sum \infty (\frac{11 π}{e})^{n}

n = 1 \sum \infty n^{- 0.9999}

n = 0 \sum \infty \frac{cos ( πn )}{n}

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