4

I try to exprex the following doble summ $$\sum _{k=0}^n \sum _{m=k+1}^n \frac{a_k \left(2^{2 m} n \binom{m+n}{2 m}\right)}{m+n}$$ it comes from acceleration series and i need to express in this way $$\sum _{j=1}^n \sum _{k=0}^{j-1} \frac{2^{2 (k+1)} n a_k \binom{k+n+1}{2 (k+1)}}{k+n+1}$$ I try to eleminate the m index I do not what is it wrong but the series are different I preciate any help in this sense Thanks

capea
  • 456
  • Do you claim the two things should be equal? Because they are not, for $n=2$ you have $6a_0+32a_0+32a_1\neq6a_0+6a_0+8a_1$, if I computed it correctly. What exactly do you want? – yo' Oct 28 '13 at 21:24
  • the two series ,you wright are not equal, the second if my tries to transform the series and eliminate the m index but i could do it – capea Oct 28 '13 at 21:36

2 Answers2

2

You have $$\sum_{k=0}^n \sum_{m=k+1}^n a_k b_m,$$ what is $b_m$ is not interesting now. This all means that you have $$0\leq k \leq m-1 \leq n-1,\quad\text{i.e.},\quad 1\leq k+1\leq m\leq n.$$ The proper sum exchange is therefore $$\sum_{m=1}^n \sum_{k=0}^{m-1} b_m a_k.$$

For me, writing all four important variables (lower criterion, upper criterion and both summation variables) in one inequality always solves the sum exchange and I've never made a mistake this way.

Btw, notice that for $k=n$ the sum is empty.

yo'
  • 4,506
2

You have $$\sum_{k=0}^n\sum_{m=k+1}^n=\sum_{k,m:}_{0\le k<m\le n}=\sum_{m=1}^n\sum_{k=0}^{m-1}$$

Then you can replace $m$ by $j$ if you like, but the two series you give are not equal.

vadim123
  • 82,796