Key to "Find ∑n=0∞2n32n−1+1"

kuntungw3

Answered question

2022-01-21

Find

\sum_{n = 0}^{\infty} \frac{2^{n}}{3^{2^{n} - 1} + 1}

trnovitom06

Beginner2022-01-22Added 12 answers

Start with following infinite product expansion which is valid for

| q | < 1

\prod_{n = 0}^{\infty} (1 + q^{2^{n}}) = \sum_{n = 0}^{\infty} q^{n} = \frac{1}{1 - q}

Taking logarithm and apply

q \frac{d}{d q}

on both sides, one get

\sum_{n = 0}^{\infty} \frac{2^{n} q^{2^{n}}}{1 + q^{2^{n}}} = \frac{q}{1 - q}

Substitute q by

\frac{1}{\sqrt{3}}

, one obtain

\sum_{n = 0}^{\infty} \frac{2^{n}}{3^{2^{n - 1}} + 1} = \frac{1}{\sqrt{3} - 1} = \frac{\sqrt{3} + 1}{2} = 1 + \frac{\sqrt{3} - 1}{2} = 1 + \frac{1}{\sqrt{3} + 1}

the expression you expected.

ocretz56

Beginner2022-01-23Added 16 answers

Theres

RizerMix

Expert2022-01-27Added 656 answers

There is no reason for the oddball

\frac{2^{0}}{3^{2^{0 - 1}}}

term so I omit it.

\sum_{r = 1}^{\infty} \frac{2^{r}}{3^{2^{r - 1}}} \frac{2^{r}}{3^{2^{r - 1}}} = \sum_{r = 1}^{\infty} 2^{r} \sum_{m = 1}^{\infty} (- 1)^{m - 1} 3^{- 2^{r - 1} m}

= \sum_{n = 1}^{\infty} 3^{- n} \sum_{2^{r - 1} | n} 2^{r} (- 1)^{n / 2^{r} - 1}

= \sum_{n = 1}^{\infty} 3^{- n} \cdot 2

= 2 \frac{1 / 3}{1 - 1 / 3}

= 1

Note, given

v := v_{2} (n), \sum_{2^{r - 1} | n} 2^{r} (- 1)^{n / 2^{r} - 1}

is 2 if

v = 0

(i.e. n is odd) and otherwise

= - 2 \sum_{r = 0}^{v} 2^{r} (- 1)^{n / 2^{r}} = - 2 [\sum_{r = 0}^{v - 1} 2^{r} - 2^{v}]

= - 2 [\frac{2^{v} - 1}{2 - 1} - 2^{v}] = - 2 (- 1) = 2

v > 0

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