Find-the-value-of-i-0-j-0-k-0-1-3-i-3-j-3-k-case-1-i-j-k-case-2-i-lt-j-lt-k-

Question Number 31743 by rahul 19 last updated on 13/Mar/18

F i n d t h e v a l u e o f : \underset{i = 0}{\sum^{\infty}} \underset{j = 0}{\sum^{\infty}} \underset{k = 0}{\sum^{\infty}} \frac{1}{3^{i} 3^{j} 3^{k}} .

c a s e 1 : i \neq j \neq k .

c a s e 2 : i < j < k .

Answered by mrW2 last updated on 13/Mar/18

\underset{k = 0}{\sum^{\infty}} \frac{1}{3^{k}} = 1 + \frac{1}{3} + \frac{1}{3^{2}} + \dots . = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2}

\underset{i = 0}{\sum^{\infty}} \underset{j = 0}{\sum^{\infty}} \underset{k = 0}{\sum^{\infty}} \frac{1}{3^{i} 3^{j} 3^{k}}

= \underset{i = 0}{\sum^{\infty}} \underset{j = 0}{\sum^{\infty}} {\frac{1}{3^{i} 3^{j}} (\underset{k = 0}{\sum^{\infty}} \frac{1}{3^{k}})}

= \frac{3}{2} \underset{i = 0}{\sum^{\infty}} \underset{j = 0}{\sum^{\infty}} \frac{1}{3^{i} 3^{j}}

= \frac{3}{2} \underset{i = 0}{\sum^{\infty}} {\frac{1}{3^{i}} (\underset{j = 0}{\sum^{\infty}} \frac{1}{3^{j}})}

= \frac{3}{2} \times \frac{3}{2} \underset{i = 0}{\sum^{\infty}} \frac{1}{3^{i}}

= \frac{3}{2} \times \frac{3}{2} \times \frac{3}{2}

= \frac{27}{8}

Commented by mrW2 last updated on 14/Mar/18

\underset{i = 0}{\sum^{\infty}} \underset{j = 0}{\sum^{\infty}} \underset{k = 0}{\sum^{\infty}} \frac{1}{3^{i} 3^{j} 3^{k}} = \frac{27}{8}

t h i s i s c o r r e c t .

i f y o u n e e d \underset{i \neq j \neq k}{\underset{i = 0}{\sum^{\infty}} \underset{j = 0}{\sum^{\infty}} \underset{k = 0}{\sum^{\infty}}} \frac{1}{3^{i} 3^{j} 3^{k}} o r

\underset{i > j > k}{\underset{i = 0}{\sum^{\infty}} \underset{j = 0}{\sum^{\infty}} \underset{k = 0}{\sum^{\infty}}} \frac{1}{3^{i} 3^{j} 3^{k}} t h e n t h e r e s u l t i s d i f f e r e n t .

Commented by rahul 19 last updated on 14/Mar/18

B u t w h a t i s t h e d i f f e r e n c e w h e n

i \neq j \neq k a n d i < j < k a r e p r e s e n t i n q u e s .

(o r i g i n a l q .) a n d w h e n t h a t i s n o t m e n t i o n

i n t h e q u e s . (t h e w a y u s o l v e d) .

Commented by mrW2 last updated on 14/Mar/18

l o o k a t a n e x a m p l e :

\underset{i = 1}{\sum^{3}} \underset{j = 1}{\sum^{3}} a^{i} b^{j} = a^{1} (b^{1} + b^{2} + b^{3}) + a^{2} (b^{1} + b^{2} + b^{3}) + a^{3} (b^{1} + b^{2} + b^{3}) = (a^{1} + a^{2} + a^{3}) (b^{1} + b^{2} + b^{3})

(i a n d j a r e i n d e p e n d e n t)

\underset{i \neq j}{\underset{i = 1}{\sum^{3}} \underset{j = 1}{\sum^{3}}} a^{i} b^{j} = a^{1} (b^{2} + b^{3}) + a^{2} (b^{1} + b^{3}) + a^{3} (b^{1} + b^{2})

(i a n d j a r e d e p e n d e n t)

\underset{i < j}{\underset{i = 1}{\sum^{3}} \underset{j = 1}{\sum^{3}}} a^{i} b^{j} = a^{1} (b^{2} + b^{3}) + a^{2} b^{3}

(i a n d j a r e d e p e n d e n t)

Commented by rahul 19 last updated on 14/Mar/18

o k s i r, n o w u n d e r s t o o d w h e n i t s i n

2 v a r i a b l e s (i, j) .

p l s a l s o s h o w s i m i l a r e x a m p l e f o r

3 v a r i a b l e (i, j, k) ?

Commented by abdo imad last updated on 14/Mar/18

t h e r e i s a d f f e r e n c e b e t w e e n \sum_{i j} a_{i} b j a n d \sum_{i, j a n d i \neq j} a_{i} a_{j}

a n d t h e t y p e o f t h i s t r i p l e s u m i s d i f f i c u t t o t r e a t

a n d i f y o u g i v e i n t e r e s t y o u m u s t t a k e a l o o k t o

s o m m a b l e s f a m i l y \dots .