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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply p-Series Test:

converges

= \frac{5}{2} converges

= converges

n = 1 \sum \infty 7 - 2 (0.9)^{n}

n = 1 \sum \infty (\frac{4}{25})^{n - 1}

n = 1 \sum \infty \frac{8}{4 n ^{2} - 1}

n = 0 \sum \infty \frac{2 ^{n}}{n ^{5}}

n = 1 \sum \infty cos^{n} (15)

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