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Show that this identity; $$\sum_{i=1}^{n}\sum_{j=1}^{n}\min{\{i,j\}}(a_{i}-a_{i+1})(b_{j}-b_{j+1})=\sum_{i=1}^{n}a_{i}b_{i}$$ where $a_{i},b_{i}\in R,(i=1,2,\cdots,n),a_{n+1}=b_{n+1}=0$

It seem can use integral to show it? And maybe other methods also can solve it?

Martin Sleziak
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math110
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    Seems unlikely as $a_{n+1},b_{n+1}$ do not appear in the RHS – Bart Michels Jun 23 '17 at 12:06
  • You can't prove it. It's simply not true. Take n= 2, $a_1= 1$, $a_2= 3$, $a_3= 6$, $b_1= 8$, $b_2= 4$, $b_3= 3$. Then the left side is tex(1- 3)(8- 4)+ (1)(1- 3)(4- 3)+ (1)(6- 3)(8- 4)+ (2)(6- 3)(1- 3)= (-2)(4)- (2)(1)+ 3(4)+ (2)3(-2)= -8- 2+ 12- 12= -10. The right side is 1(8)+ 3(4)+ 6(3)= 8+ 12+ 18= 38. – user247327 Jun 23 '17 at 12:15
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    @user247327 For $n=2$, you should have $a_3=b_3=0$, as stated in the question. – egreg Jun 23 '17 at 12:17

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Split into two cases: $$ \sum_{i=1}^n \sum_{j=1}^i j(a_i-a_{i+1})(b_j-b_{j+1}) + \sum_{j=2}^n \sum_{i=1}^{j-1} i(a_i-a_{i+1})(b_j-b_{j+1}) \\ = \sum_{i=1}^n (a_i-a_{i+1}) \sum_{j=1}^i j(b_j-b_{j+1}) + \sum_{j=2}^n (b_j-b_{j+1}) \sum_{i=1}^{j-1} i(a_i-a_{i+1}) $$ (note the sums are in a different order in each term). Consider the inner sum in the first term: $$ \sum_{j=1}^i j(b_j-b_{j+1}) = \sum_{j=1}^i \sum_{k=1}^j (b_j-b_{j+1}) = \sum_{k=1}^i \sum_{j=k}^i (b_j-b_{j+1}), $$ introducing a new sum and changing the order of summation. But the inner sum telescopes, to $b_k-b_{i+1}$, so the first sum is now $$ \sum_{i=1}^n (a_i-a_{i+1}) \sum_{k=1}^i (b_k-b_{i+1}) = \sum_{i=1}^n (a_i-a_{i+1}) \sum_{k=1}^i b_k - \sum_{i=1}^n i(a_i-a_{i+1})b_{i+1} \\ =\sum_{k=1}^n b_k \sum_{i=k}^n (a_i-a_{i+1}) - \sum_{i=1}^{n-1} i(a_i-a_{i+1})b_{i+1} $$ The first sum telescopes again, to $\sum_{k=1}^n b_k (a_k-a_{n+1}) = \sum_{k=1}^n b_k a_k$ since $a_{n+1}=0$.

We apply a similar procedure to the second sum: $$ \sum_{i=1}^{j-1} i(a_i-a_{i+1}) = \sum_{i=1}^{j-1}\sum_{k=1}^i (a_i-a_{i+1}) = \sum_{k=1}^{j-1} \sum_{i=k}^{j-1} (a_i-a_{i+1}) = \sum_{k=1}^{j-1} (a_k-a_j), $$ so the second sum becomes $$ \sum_{j=2}^n (b_j-b_{j+1}) \sum_{k=1}^{j-1} (a_k-a_j) = \sum_{j=2}^n (b_j-b_{j+1}) \sum_{k=1}^{j-1} a_k - \sum_{j=2}^n (j-1)(b_j-b_{j+1})a_j \\ = \sum_{k=1}^{n-1} a_k \sum_{j=k+1}^{n} (b_j-b_{j+1}) - \sum_{j=2}^n (j-1)a_j(b_j-b_{j+1}), $$ and the first sum telescopes to $ \sum_{k=1}^{n-1}a_k b_{k+1} $. So we are left with the total sum being equal to $$ \sum_{k=1}^n b_k a_k + \sum_{k=1}^{n-1}a_k b_{k+1} - \sum_{k=1}^{n-1} k(a_k-a_{k+1})b_{k+1} - \sum_{k=1}^n (k-1)a_k(b_k-b_{k+1}), $$ re-indexing the sums. The last three terms are $$ \sum_{k=1}^{n-1}a_k b_{k+1} - \sum_{k=1}^{n-1} ka_k b_{k+1} - \sum_{k=1}^{n-1} k a_{k+1} b_{k+1} - \sum_{k=2}^n (k-1)a_k b_k + \sum_{k=1}^{n-1} (k-1)a_k b_{k+1} =0, $$ and the result follows.

Chappers
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