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

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

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

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

converges

Solution steps

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

Simplify

(\frac{1}{n})^{2} : \frac{1}{n ^{2}}

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

Apply p-Series Test:

converges

= converges

n = 0 \sum \infty cos (8 n)

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

n = 1 \sum \infty n^{e - π}

n = 1 \sum \infty 8^{\frac{1}{n}}

n = 1 \sum \infty \frac{n - 1}{4 n - 1}

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