calculate-S-n-0-i-j-n-i-j-2-i-j-

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

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}} + \dots + \frac{n}{2^{n}}

2 T = 1 + \frac{1 + 1}{2} + \frac{1 + 2}{2^{2}} + \frac{1 + 3}{2^{3}} \dots + \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}}) \dots + (\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}} + \dots + \frac{1}{2^{n - 1}}) + (\frac{1}{2} + \frac{2}{2^{2}} + \frac{3}{2^{3}} \dots + \frac{n - 1}{2^{n - 1}})

2 T = (1 + \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \dots + \frac{1}{2^{n - 1}}) + (\frac{1}{2} + \frac{2}{2^{2}} + \frac{3}{2^{3}} \dots + \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}} + \dots + \frac{1}{2^{n - 1}}) + T - \frac{n}{2^{n}}

T = (1 + \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \dots + \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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