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

The sum from k=1 to infinity of 1/(k^2) is converges

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

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

converges

Solution steps

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

Apply p-Series Test:

converges

= converges

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

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

n = 1 \sum \infty (\frac{99}{100})^{n}

n = 1 \sum \infty \frac{7}{n ^{3}}

n = 1 \sum \infty 3 (\frac{4}{5})^{n}

What is the sum from k=1 to infinity of 1/(k^2) ?
The sum from k=1 to infinity of 1/(k^2) is converges