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

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

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

n = 1 \sum \infty (\frac{2}{n})^{2}

converges

Solution steps

n = 1 \sum \infty (\frac{2}{n})^{2}

Simplify

(\frac{2}{n})^{2} : \frac{4}{n ^{2}}

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

Apply the constant multiplication rule:

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

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

Apply p-Series Test:

converges

= 4 converges

= converges

n = 1 \sum \infty (\frac{2}{n})^{1}

n = 0 \sum \infty 3 + \frac{2}{n ^{4}}

n = 1 \sum \infty \frac{3}{n ^{9}}

n = 0 \sum \infty \frac{n ^{2}}{1 + 6 n}

n = 1 \sum \infty \frac{2}{n ^{9}}

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