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

The sum from n=1 to infinity of 1/(2^{2n+1)} is 1/6

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

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

\frac{1}{6}

Solution steps

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

Simplify

2^{2 n + 1} : 2^{2 n} \cdot 2

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

Apply the constant multiplication rule:

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

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

2^{2 n} = 4^{n}

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

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= \frac{1}{2} (n = 0 \sum \infty \frac{1}{4 ^{n}} - \frac{1}{4 ^{0}})

n = 0 \sum \infty \frac{1}{4 ^{n}} = \frac{4}{3}

= \frac{1}{2} (\frac{4}{3} - \frac{1}{4 ^{0}})

Simplify

\frac{1}{2} (\frac{4}{3} - \frac{1}{4 ^{0}}) : \frac{1}{6}

= \frac{1}{6}

n = 2 \sum \infty \frac{6}{3 ^{2 n}}

n = 0 \sum \infty \frac{ln ( n ^{2} )}{n}

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

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

n = 1 \sum \infty ln (\frac{1}{3 ^{n}})

What is the sum from n=1 to infinity of 1/(2^{2n+1)} ?
The sum from n=1 to infinity of 1/(2^{2n+1)} is 1/6