What is the sum from n=0 to infinity of 3/(n^2) ?

The sum from n=0 to infinity of 3/(n^2) is converges

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sum from n=0 to infinity of 3/(n^2)

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

converges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply p-Series Test:

converges

= 3 converges

= converges

n = 1 \sum \infty (1 - \frac{6}{n})^{n}

k = 2 \sum \infty k e^{- 4 k^{2}}

n = 1 \sum \infty \frac{1 0 ^{n}}{n}

n = 0 \sum \infty 0. 6^{n}

n = 0 \sum \infty cos (\frac{1}{n ^{3}})

What is the sum from n=0 to infinity of 3/(n^2) ?
The sum from n=0 to infinity of 3/(n^2) is converges