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

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

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

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

converges

Solution steps

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

Simplify

\frac{1}{( 2 n ) ^{2}} : \frac{1}{4 n ^{2}}

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

Apply the constant multiplication rule:

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

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

Apply p-Series Test:

converges

= \frac{1}{4} converges

= converges

n = 1 \sum \infty n^{2} (\frac{5}{8})^{n}

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

k = 0 \sum \infty (- \frac{5}{6})^{k}

n = 1 \sum \infty (\frac{1}{100})^{n}

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

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