calculate S_n =Σ_(0≤i,j≤n) ((i+j)/2^(i+j) )


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Question Number 46744 by math khazana by abdo last updated on 31/Oct/18

$c a l c u l a t e S_{n} = \sum_{0 ⩽ i, j ⩽ n} \frac{i + j}{2^{i + j}}$
Commented by MJS last updated on 31/Oct/18

$I think S_{\infty} = 8 but I cannot prove it$
Commented by maxmathsup by imad last updated on 31/Oct/18
$we have S_n =Σ_(i=0) ^n Σ_(j=0) ^n ((i/(2^i 2^j )) +(j/(2^i 2^j ))) =Σ_(i=0) ^n (i/2^i )Σ_(j=0) ^n (1/2^j ) +Σ_(i=0) ^n (1/2^i )Σ_(j=0) ^n (j/2^j ) = 2 Σ_(i=0) ^n (i/2^i )Σ_(j=0) ^n (1/2^j )( the function (i,j)→((i+j)/2^(i+j) ) is symetric) we have Σ_(j=0) ^n (1/2^j ) =((1−((1/2))^(n+1) )/(1−(1/2))) =2(1−(1/2^(n+1) )) =2 −(1/2^n ) also Σ_(i=0) ^n (i/2^i ) = w((1/2)) with w(x)=Σ_(i=0) ^n ix^i (x≠1) we have Σ_(i=0) ^n x^i =((x^(n+1) −1)/(x−1)) ⇒Σ_(i=1) ^n i x^(i−1) =((nx^(n+1) −(n+1)x^n +1)/((1−x)^2 )) ⇒ Σ_(i=1) ^n i x^i = (x/((1−x)^2 )){nx^(n+1) −(n+1x^n +1}=w(x) ⇒ w((1/2)) =(1/(2((1/4)))){(n/2^(n+1) ) −((n+1)/2^n ) +1} =(n/2^n ) −((n+1)/2^(n−1) ) +2 ⇒ S_n =2 ( (n/2^n ) −((n+1)/2^(n−1) ) +2)(2−(1/2^n )) and we see that lim_(n→+∞) S_(n ) =8 .$
$w e h a v e S_{n} = \sum_{i = 0}^{n} \sum_{j = 0}^{n} (\frac{i}{2^{i} 2^{j}} + \frac{j}{2^{i} 2^{j}})$ $= \sum_{i = 0}^{n} \frac{i}{2^{i}} \sum_{j = 0}^{n} \frac{1}{2^{j}} + \sum_{i = 0}^{n} \frac{1}{2^{i}} \sum_{j = 0}^{n} \frac{j}{2^{j}} = 2 \sum_{i = 0}^{n} \frac{i}{2^{i}} \sum_{j = 0}^{n} \frac{1}{2^{j}} (t h e f u n c t i o n$ $(i, j) \to \frac{i + j}{2^{i + j}} i s s y m e t r i c) w e h a v e \sum_{j = 0}^{n} \frac{1}{2^{j}} = \frac{1 - {(\frac{1}{2})}^{n + 1}}{1 - \frac{1}{2}} = 2 (1 - \frac{1}{2^{n + 1}})$ $= 2 - \frac{1}{2^{n}} a l s o \sum_{i = 0}^{n} \frac{i}{2^{i}} = w (\frac{1}{2}) w i t h w (x) = \sum_{i = 0}^{n} {i x}^{i} (x \neq 1)$ $w e h a v e \sum_{i = 0}^{n} x^{i} = \frac{x^{n + 1} - 1}{x - 1} \Rightarrow \sum_{i = 1}^{n} i x^{i - 1} = \frac{{n x}^{n + 1} - (n + 1) x^{n} + 1}{{(1 - x)}^{2}} \Rightarrow$ $\sum_{i = 1}^{n} i x^{i} = \frac{x}{{(1 - x)}^{2}} {{n x}^{n + 1} - (n + 1 x^{n} + 1} = w (x) \Rightarrow$ $w (\frac{1}{2}) = \frac{1}{2 (\frac{1}{4})} {\frac{n}{2^{n + 1}} - \frac{n + 1}{2^{n}} + 1} = \frac{n}{2^{n}} - \frac{n + 1}{2^{n - 1}} + 2 \Rightarrow$ $S_{n} = 2 (\frac{n}{2^{n}} - \frac{n + 1}{2^{n - 1}} + 2) (2 - \frac{1}{2^{n}}) a n d w e s e e t h a t {l i m}_{n \to + \infty} S_{n} = 8 .$
	Answered by MrW3 last updated on 31/Oct/18

	$\frac{i + j}{2^{i + j}} = \frac{i + j}{2^{i} 2^{j}} = \frac{i}{2^{i}} \times \frac{1}{2^{j}} + \frac{1}{2^{i}} \times \frac{j}{2^{j}}$ $S_{n} = \underset{i = 0}{\sum^{n}} \underset{j = 0}{\sum^{n}} \frac{i + j}{2^{i + j}} = \underset{i = 0}{\sum^{n}} \frac{i}{2^{i}} \underset{j = 0}{\sum^{n}} \frac{1}{2^{j}} + \underset{i = 0}{\sum^{n}} \frac{1}{2^{i}} \underset{j = 0}{\sum^{n}} \frac{j}{2^{j}}$ $\underset{i = 0}{\sum^{n}} \frac{1}{2^{i}} = \frac{1 - \frac{1}{2^{n + 1}}}{1 - \frac{1}{2}} = 2 (1 - \frac{1}{2^{n + 1}})$ $l e t T = \underset{i = 0}{\sum^{n}} \frac{i}{2^{i}}$ $T = \frac{1}{2} + \frac{2}{2^{2}} + \frac{3}{2^{3}} + . . . + \frac{n}{2^{n}}$ $2 T = 1 + \frac{1 + 1}{2} + \frac{1 + 2}{2^{2}} + \frac{1 + 3}{2^{3}} . . . + \frac{1 + n - 1}{2^{n - 1}}$ $2 T = 1 + (\frac{1}{2} + \frac{1}{2}) + (\frac{1}{2^{2}} + \frac{2}{2^{2}}) + (\frac{1}{2^{3}} + \frac{3}{2^{3}}) . . . + (\frac{1}{2^{n - 1}} + \frac{n - 1}{2^{n - 1}})$ $2 T = (1 + \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + . . . + \frac{1}{2^{n - 1}}) + (\frac{1}{2} + \frac{2}{2^{2}} + \frac{3}{2^{3}} . . . + \frac{n - 1}{2^{n - 1}})$ $2 T = (1 + \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + . . . + \frac{1}{2^{n - 1}}) + (\frac{1}{2} + \frac{2}{2^{2}} + \frac{3}{2^{3}} . . . + \frac{n - 1}{2^{n - 1}} + \frac{n}{2^{n}}) - \frac{n}{2^{n}}$ $2 T = (1 + \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + . . . + \frac{1}{2^{n - 1}}) + T - \frac{n}{2^{n}}$ $T = (1 + \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + . . . + \frac{1}{2^{n - 1}}) - \frac{n}{2^{n}}$ $T = 2 (1 - \frac{1}{2^{n}}) - \frac{n}{2^{n}} = 2 - \frac{n + 2}{2^{n}} = 2 (1 - \frac{n + 2}{2^{n + 1}})$ $S_{n} = \underset{i = 0}{\sum^{n}} \underset{j = 0}{\sum^{n}} \frac{i + j}{2^{i + j}} = \underset{i = 0}{\sum^{n}} \frac{i}{2^{i}} \underset{j = 0}{\sum^{n}} \frac{1}{2^{j}} + \underset{i = 0}{\sum^{n}} \frac{1}{2^{i}} \underset{j = 0}{\sum^{n}} \frac{j}{2^{j}}$ $= 2 \underset{i = 0}{\sum^{n}} \frac{i}{2^{i}} \underset{i = 0}{\sum^{n}} \frac{1}{2^{i}} = 2 \times 2 (1 - \frac{n + 2}{2^{n + 1}}) \times 2 (1 - \frac{1}{2^{n + 1}})$ $\Rightarrow S_{n} = 8 (1 - \frac{n + 2}{2^{n + 1}}) (1 - \frac{1}{2^{n + 1}})$ $\lim_{n \to \infty} S_{n} = 8$
	Commented by behi83417@gmail.com last updated on 31/Oct/18

	$e X \subseteq e l l e^{N t}!$
	Commented by maxmathsup by imad last updated on 31/Oct/18

	$t h a n k y o u s i r .$
	Commented by MJS last updated on 31/Oct/18

	$great$
	Commented by MrW3 last updated on 31/Oct/18

	$t h a n k s t o a l l!$ $p l e a s e a l s o c h e c k a n s w e r t o Q 46731 .$
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