What is the sum from k=1 to infinity of k^2 ?

The sum from k=1 to infinity of k^2 is diverges

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sum from k=1 to infinity of k^2

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

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

n = 1 \sum \infty 5 (\frac{1}{5})^{n - 1}

n = 1 \sum \infty \frac{ln ( n )}{9 n}

n = 2 \sum \infty \frac{2}{7 ^{n}}

n = 1 \sum \infty \frac{ln ( n )}{3 n}

What is the sum from k=1 to infinity of k^2 ?
The sum from k=1 to infinity of k^2 is diverges