What is the sum from n=1 to infinity of 5/(3n^2) ?

The sum from n=1 to infinity of 5/(3n^2) is converges

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sum from n=1 to infinity of 5/(3n^2)

n = 1 \sum \infty \frac{5}{3 n ^{2}}

converges

Solution steps

n = 1 \sum \infty \frac{5}{3 n ^{2}}

Apply the constant multiplication rule:

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

= \frac{5}{3} \cdot n = 1 \sum \infty \frac{1}{n ^{2}}

Apply p-Series Test:

converges

= \frac{5}{3} converges

= converges

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

n = 1 \sum \infty \frac{1}{3 n ^{3} + 8}

n = 1 \sum \infty e^{- \frac{n}{9}}

n = 1 \sum \infty \frac{1}{n ^{10}}

n = 1 \sum \infty cos (\frac{nπ}{2})

What is the sum from n=1 to infinity of 5/(3n^2) ?
The sum from n=1 to infinity of 5/(3n^2) is converges