What is the sum from n=0 to infinity of 1/(3n^2) ?

The sum from n=0 to infinity of 1/(3n^2) is converges

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sum from n=0 to infinity of 1/(3n^2)

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply p-Series Test:

converges

= \frac{1}{3} converges

= converges

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

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

n = 1 \sum \infty \frac{1}{n ^{7} + 2}

k = 0 \sum \infty \frac{k}{1 + k ^{4}}

n = 1 \sum \infty \frac{5}{n ^{0.6}}

What is the sum from n=0 to infinity of 1/(3n^2) ?
The sum from n=0 to infinity of 1/(3n^2) is converges