47105 $2^n$在分母的級數收斂性質補遺
$2^n$在分母的級數收斂性質補遺

$$\sum_{n=1}^\infty \frac{L_n(a,b)}{2^n}=a+b, \ \sum_{n=1}^\infty \frac{L_n(a,b,c)}{2^n}=a+b+c,\ \hbox{甚至}\ \sum_{n=1}^\infty\frac{L_n(a,b,c,d)}{2^n}=a+b+c+d,$$

1. 降階公式 ： $d_i=\sum\limits_{j=0}^i(-1)^j a_{i+1-j}C_j^i$, (由 $a_n$ 求 $d_i$)。

2. 生成公式 ： $a_n=\sum\limits_{j=0}^r d_j C_j^{n-1}$, (由 $d_i$ 求 $a_n$)。

3. 級數和公式： $S_n=\sum\limits_{j=0}^r d_j C_{j+1}^n$, (由 $d_i$ 求 $S_n$)。

\begin{align*} a_n=&a\cdot C_0^{n-1}+b\cdot C_1^{n-1}+c\cdot C_2^{n-1}\\ =&a+b(n-1)+c \frac{(n-1)(n-2)}{2!}\\ =&\frac c 2 n^2+\frac{2b-3c}2 n+a-b+c,\\ {\hbox{所以}} \sum_{n=1}^\infty \frac{a_n}{2^n}=& \sum_{n=1}^\infty \frac{[a,b,c]}{2^n}=\frac c2\cdot \sum_{n=1}^\infty \frac{n^2}{2^n}+\frac{2b-3c}{2}\cdot \sum_{n=1}^\infty \frac{n}{2^n}+\sum_{n=1}^\infty \frac{a-b+c}{2^n}\\ =&\frac c2\cdot 6+\frac{2b-3c}2\cdot 2+a-b+c\\ =&a+b+c.\tag*{得證。} \end{align*}

$$\{a_n\}=[a,b,c,d] \quad \hbox{時},\quad \sum_{n=1}^\infty \frac{a_n}{2^n}=a+b+c+d \hbox{。}$$

\begin{align*} a_n=&a\cdot C_0^{n-1}+b\cdot C_1^{n-1}+c\cdot C_2^{n-1}+d\cdot C_3^{n-1},\\ {\hbox{所以}} \sum_{n=1}^\infty \frac{a_n}{2^n} =&a\cdot \sum_{n=1}^\infty \frac{C_0^{n-1}}{2^n}+b\cdot \sum_{n=1}^\infty \frac{C_1^{n-1}}{2^n} +c\cdot \sum_{n=1}^\infty \frac{C_2^{n-1}}{2^n}+d\cdot \sum_{n=1}^\infty \frac{C_3^{n-1}}{2^n}\\ =&a\cdot \sum_{n=1}^\infty \frac 1{2^n} +\frac b2\cdot\sum_{n=1}^\infty \frac{C_1^n}{2^n}+\frac c4\cdot \sum_{n=1}^\infty \frac{C_2^{n+1}}{2^n} +\frac d8\cdot \sum_{n=1}^\infty \frac{C_3^{n+2}}{2^n}\tag*{(*)}\\ =&a +\frac b2\cdot \sum_{n=1}^\infty \frac n{2^n}+\frac c4\cdot \sum_{n=1}^\infty \frac{(n+1)n}{{2^n}\cdot 2!}+\frac d8\cdot \sum_{n=1}^\infty \frac{(n+2)(n+1)n}{2^n\cdot 3!}\\ =&a+\frac b2\cdot 2+\frac c8\cdot \sum_{n=1}^\infty \frac{n^2+n}{2^n}+\frac d{48}\cdot \sum_{n=1}^\infty \frac{n^3+3n^2+2n}{2^n}\\ =&a+b+\frac c8\cdot (6+2)+\frac{d}{48}\cdot (26+18+4)\\ =&a+b+c+d.\tag*{得證。} \end{align*}

(在 $(*)$ 的式子中, 要扣除當 $n\lt r$ 時, $C_r^n=0$ 的各項。)

$$\sum\limits_{n=1}^\infty \dfrac{n^4}{2^n} =\sum\limits_{n=1}^\infty \dfrac{[1,15,50,60,24]}{2^n}=1+15+50+60+24=150,$$

$$\sum\limits_{n=1}^\infty \dfrac{S_n}{2^{n+1}} =\sum\limits_{n=1}^\infty \dfrac{S'_n}{2^n}=\sum\limits_{n=1}^\infty \dfrac{[0,d_0,d_1,\ldots,d_r ]}{2^n} =0+d_0+d_1+\cdots+d_r=\sum\limits_{n=1}^\infty \dfrac{a_n}{2^n}\ \hbox{。}$$

$$\sum_{n=1}^\infty \frac{S_n}{2^{n+1}} =\sum\limits_{n=1}^\infty \frac{a_n}{2^n}=\sum_{j=0}^r d_j\ \hbox{。}$$

$$\sum_{n=1}^\infty \frac{a_n}{2^n} =\sum_{n=1}^\infty \frac{\sum_{j=0}^r d_j C_j^{n-1}}{2^n} =\sum\limits_{j=0}^r d_j\Big(\sum_{n=1}^\infty \frac{C_j^{n-1}}{2^n}\Big),$$

\begin{align*} &\sum_{n=1}^\infty \frac c{2^n} =c\\ &\sum_{n=1}^\infty \frac{a_1+(n-1)d}{2^n} =\sum_{n=1}^\infty \frac{[a_1,d]}{2^n}=a_1+d\\ &\sum_{n=1}^\infty \frac{F_n}{2^n}=\sum_{n=1}^\infty \frac{L_n(1,1)}{2^n}=1+1\\ &\sum_{n=1}^\infty \frac{L_n (a,b,c,\ldots,k)}{2^n}=a+b+c+\cdots+k\qquad\hbox{及}\\ &\sum_{n=1}^\infty \frac{[d_0,d_1,d_2,\ldots,d_r ]}{2^n} =d_0+d_1+d_2+\cdots+d_r \quad \hbox{都是這樣的。}\hskip 2.5cm~ \end{align*}

### 後記與待解問題

$1,1,2,2,3,3,4,4,5,5,6,6,\ldots$ 的數列, 或 $1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,\ldots$ 的數列,

$$\sum_{n=1}^\infty \frac{N_2 (n)}{2^n} =\sum_{n=1}^\infty \dfrac{\Big[\frac{2-1+n}2\Big]}{2^n} =\frac 43 ,\quad \hbox{及}\ \sum_{n=1}^\infty \frac{N_3 (n)}{2^n} =\sum_{n=1}^\infty \dfrac{\Big[\frac{3-1+n}3\Big]}{2^n}=\frac{8}7 ；$$

$$\sum_{n=1}^\infty \dfrac{N_k (n)}{2^n} =\sum_{n=1}^\infty \dfrac{\Big[\frac{k-1+n}k\Big]}{2^n} =\dfrac{2^k}{2^k-1}.$$

$\{n\}$ 的自然數列 : $1,2,3,4,5,6,7,8,9,10,11,12,13,14, \ldots$,

$\{N_N (n)\}$ 數列 : $1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,6, \ldots$,

$\{N_2 (n)\}$ 數列 : $1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8, \ldots$,

$\{N_3 (n)\}$ 數列 : $1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6, \ldots$,

$$\sum\limits_{n=1}^\infty \dfrac 1{2^{((n-1)n)/2}} \fallingdotseq %\approx %≒ 1.6416325606551538\cdots.$$