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 (sin (3))^{n}

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

n = 0 \sum \infty \frac{15}{6 ^{n}}

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

n = 0 \sum \infty (- 1)^{n} \cdot 1^{n}

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