Prove-m-0-n-0-mn-2-m-n-4-

Question Number 3735 by prakash jain last updated on 19/Dec/15

Prove

\underset{m = 0}{\sum^{\infty}} \underset{n = 0}{\sum^{\infty}} \frac{m n}{2^{m + n}} = 4

Commented by prakash jain last updated on 19/Dec/15

As a part answer to Q 3716 .

Answered by prakash jain last updated on 19/Dec/15

\underset{m = 0}{\sum^{\infty}} \frac{m}{2^{m}} \underset{n = 0}{\sum^{\infty}} \frac{n}{2^{n}} = \underset{m = 0}{\sum^{\infty}} \frac{2 m}{2^{m}} = 4

Q 3734 .

Commented by Yozzii last updated on 19/Dec/15

W h a t n e c e s s a r y a n d s u f f i c i e n t

c o n d i t i o n s e x i s t s o t h a t

\sum_{m \in N} \sum_{n \in N} f (m) g (n) = {\sum_{m \in N} f (m)} {\sum_{n \in N} f (n)}

w i t h f, g > 0 ?

S = \sum_{m \in N} \sum_{n \in N} f (m) g (m) = \sum_{m \in N} {\sum_{n \in N} f (m) g (n)}

G i v e n t h a t m a n d a r e i n d e p e n d e n t

v a r i a b l e s, f o r e a c h m w e h a v e

\sum_{n \in N} f (m) g (n) = f (m) \times \sum_{n \in N} g (n)

∴ S = \sum_{m \in N} {f (m) \times \sum_{n \in N} g (n)}

I f \sum_{n \in N} g (n) c o n v e r g e s t o l

\Rightarrow S = \sum_{m \in N} f (m) \times l = l \sum_{m \in N} f (m)

S o, S = {\sum_{m \in N} f (m)} \times {\sum_{n \in N} g (n)} ?

Commented by prakash jain last updated on 19/Dec/15

f (m) i s c o n s t a n t f o r s u m m a t i o n o v e r n .

t a k i n g a c o n s t a n t o u t f o r s u m m a t i o n i s

a l w a y s v a l i d (e v e n i f t h e s e r i e s i s d i v e r g e n t) .

a b s o l u t e c o n v e r g e n c e i s r e q u i r e d f o r r e a r r a n g m n t

c h a n g i n g o r d e r o f s u m m a t i o n .

Commented by Yozzii last updated on 19/Dec/15

T h a n k y o u .

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