I can't prove it either, but I have a framework where this result fits. If you write the sum as $$S_n = \sum_{i,j=1}^n \frac1{2i+1}\frac1{2j+1} \frac{(-1)^{i+j}(n+1)^2}{i+j}{n\choose i}{n+i\choose i-1} {n\choose j}{n+i\choose j-1}$$ and use the fact that \(\displaystyle {n+i\choose i} = \frac{n+1}i{n+i\choose i-1}\), you see that $$S_n = \sum_{i,j=1}^n \frac1{2i+1}\frac1{2j+1} \frac{(-1)^{i+j}ij}{i+j}{n\choose i}{n+i\choose i} {n\choose j}{n+i\choose j}.$$ That sum looks remarkably similar to those that occurred in http://mathhelpboards.com/linear-abstract-algebra-14/find-a_1-3-a_2-5-a-7056.html (see in particular the equations labelled (1) and (2) in my comment #4 there). Using the notation of that thread, let $H_n$ be the $n\times n$ Hilbert matrix, let $H_n^{-1} = X_n^*X_n$ be the factorisation of its inverse into lower- and upper-triangular matrices, and let $k$ be the vector $\bigl(\frac13,\frac15,\ldots, \frac1{2n+1}\bigr).$ Then your sum $S_n$ is exactly one-quarter of the sum in that other thread, namely $S_n = \langle H_n^{-1}k,k\rangle = \|X_nk\|^2$. In that thread, anemone and I decided that this must be equal to $\frac14\!\bigl(1-\frac1{(2n+1)^2}\bigr)$, which agrees with your conjecture $S_n = \frac{n(n+1)}{(2n+1)^2}.$ But I still can't prove it!Bibubo said:I have this sum $$\left(N+1\right)^{2}\underset{j=1}{\overset{N}{\sum}}\frac{\left(-1\right)^{j}}{2j+1}\dbinom{N}{j}\dbinom{N+j}{j-1}\underset{i=1}{\overset{N}{\sum}}\frac{\left(-1\right)^{i}}{\left(2i+1\right)\left(i+j\right)}\dbinom{N}{i}\dbinom{N+i}{i-1}$$ and numerical test indicates that is equal to $$\frac{\left(N+1\right)N}{\left(2N+1\right)^{2}}$$ but I'm not able to prove it. Thank you.
Yes I see what you say... I've think to use in some way the Legendre shifted polynomials $$P_{N}\left(x\right)=\left(-1\right)^{N}\underset{i=0}{\overset{N}{\sum}}\dbinom{N}{i}\dbinom{N+i}{i}\left(-x\right)^{i}$$ because the sums are very similar, and because we have $$\int_{0}^{1}P_{m}\left(x\right)P_{n}\left(x\right)dx=\frac{\delta_{m,n}}{2n+1}$$ where $\delta_{m,n}$ is the Kronecker delta. Is it a stupid idea?Opalg said:I can't prove it either, but I have a framework where this result fits. If you write the sum as $$S_n = \sum_{i,j=1}^n \frac1{2i+1}\frac1{2j+1} \frac{(-1)^{i+j}(n+1)^2}{i+j}{n\choose i}{n+i\choose i-1} {n\choose j}{n+i\choose j-1}$$ and use the fact that \(\displaystyle {n+i\choose i} = \frac{n+1}i{n+i\choose i-1}\), you see that $$S_n = \sum_{i,j=1}^n \frac1{2i+1}\frac1{2j+1} \frac{(-1)^{i+j}ij}{i+j}{n\choose i}{n+i\choose i} {n\choose j}{n+i\choose j}.$$ That sum looks remarkably similar to those that occurred in http://mathhelpboards.com/linear-abstract-algebra-14/find-a_1-3-a_2-5-a-7056.html (see in particular the equations labelled (1) and (2) in my comment #4 there). Using the notation of that thread, let $H_n$ be the $n\times n$ Hilbert matrix, let $H_n^{-1} = X_n^*X_n$ be the factorisation of its inverse into lower- and upper-triangular matrices, and let $k$ be the vector $\bigl(\frac13,\frac15,\ldots, \frac1{2n+1}\bigr).$ Then your sum $S_n$ is exactly one-quarter of the sum in that other thread, namely $S_n = \langle H_n^{-1}k,k\rangle = \|X_nk\|^2$. In that thread, anemone and I decided that this must be equal to $\frac14\!\bigl(1-\frac1{(2n+1)^2}\bigr)$, which agrees with your conjecture $S_n = \frac{n(n+1)}{(2n+1)^2}.$ But I still can't prove it!
The comment about shifted Legendre polynomials prompted me to think about this problem again. In the http://mathhelpboards.com/linear-abstract-algebra-14/find-a_1-3-a_2-5-a-7056.html#post32971, I showed how to reduce the problem to proving the identity $$\sum_{k=1}^n \frac{(-1)^kk}{(n+k)(2k+1)}{n\choose k}{n+k\choose k} = \frac{-1}{4n^2-1} \qquad(*)$$ (see equation (4) and the following sentence in that thread). I posted this identity in the MathOverflow forum, where it received an amazingly clever proof that I can't resist repeating here.Bibubo said:I've think to use in some way the Legendre shifted polynomials $$P_{N}\left(x\right)=\left(-1\right)^{N}\underset{i=0}{\overset{N}{\sum}}\dbinom{N}{i}\dbinom{N+i}{i}\left(-x\right)^{i}$$
Grat job! I have only one question... if you put $x=1$ in your identity you obtain $$\underset{k=1}{\overset{n}{\sum}}\frac{\left(-1\right)^{k}k}{\left(n+k\right)\left(2k+1\right)\left(1+2k\right)}\,\dbinom{n}{k}\dbinom{n+k}{k}$$ and it isn't like the left side of (*) because there is a another factor $1+2k$... where am I wrong? Thank you!Opalg said:The comment about shifted Legendre polynomials prompted me to think about this problem again. In the http://mathhelpboards.com/linear-abstract-algebra-14/find-a_1-3-a_2-5-a-7056.html#post32971, I showed how to reduce the problem to proving the identity $$\sum_{k=1}^n \frac{(-1)^kk}{(n+k)(2k+1)}{n\choose k}{n+k\choose k} = \frac{-1}{4n^2-1} \qquad(*)$$ (see equation (4) and the following sentence in that thread). I posted this identity in the MathOverflow forum, where it received an amazingly clever proof that I can't resist repeating here.
The trick is to write the expression $\dfrac{(-1)^n(x-2)(x-4)\cdots(x- (2n-2))}{(x+2)(x+4)\cdots(x+2n)}$ as a sum of partial fractions. If you use the cover-up rule to find the coefficient of $\dfrac1{x+2k}$, you find that it is equal to $$ \frac{(-1)^n(-2k-2)(-2k-4) \cdots(-2k-2n+2)} {\bigl((-2k+2)(-2k+4)\cdots(-2)\bigr) \bigl(2\cdot4\cdots(-2k+2n)\bigr)}.$$ There are $n-1$ factors in the numerator, each of them a multiple of $-2$. In the denominator there are $k-1$ factors that are multiples of $-2$; and $n-k$ positive factors that are multiples of $2$. after cancelling all these $-1$s and $2$s, the coefficient of $\dfrac1{x+2k}$ becomes $$\begin{aligned}\frac{(-1)^k(k+1)(k+2)\cdots(n+k-1)}{(k-1)!(n-k)!} &= \frac{(-1)^k(n+k-1)!}{k!(k-1)!(n-k)!} \\ &= \frac{(-1)^kk}{n+k}\frac{(n+k)!n!}{(k!)^2((n-k)!n!} = \frac{(-1)^kk}{(n+k)(2k+1)}{n\choose k}{n+k\choose k}.\end{aligned}$$ Thus the partial fraction decomposition is $$ \sum_{k=1}^n \frac{(-1)^kk}{(n+k)(2k+1)(x+2k)}{n\choose k}{n+k\choose k} = \dfrac{(-1)^n(x-2)(x-4)\cdots(x- (2n-2))}{(x+2)(x+4)\cdots(x+2n)}.$$ Now put $x=1$. The left side then becomes the left side of (*). The right side becomes $\dfrac{-(2n-3)!}{(2n+1)!} = \dfrac{-1}{(2n-1)(2n+1)} = \dfrac{-1}{4n^2-1}$, just as we wanted!
It's not you that is wrong, it was me. I accidentally put an extraneous $2k+1$ into the previous line. I have now edited it out.Bibubo said:Grat job! I have only one question... if you put $x=1$ in your identity you obtain $$\underset{k=1}{\overset{n}{\sum}}\frac{\left(-1\right)^{k}k}{\left(n+k\right)\left(2k+1\right)\left(1+2k\right)}\,\dbinom{n}{k}\dbinom{n+k}{k}$$ and it isn't like the left side of (*) because there is a another factor $1+2k$... where am I wrong? Thank you!
A sum binomial is a mathematical expression that involves adding two binomials together. A binomial is a polynomial with two terms, separated by a plus or minus sign.
To prove a numerical test result using sum binomials, you can use algebraic manipulation to simplify the expression and show that it is equal to the desired result. This involves applying properties of addition and factoring to the sum binomials.
Sure, for example, to prove that (a + b)(a - b) = a^2 - b^2, we can expand the left side using the FOIL method: (a + b)(a - b) = a^2 - ab + ba - b^2. Then, we can simplify the middle terms by combining like terms: a^2 - ab + ba - b^2 = a^2 - b^2. Therefore, we have proven the numerical test result using sum binomials.
Sum binomials have many applications in fields such as statistics, economics, and engineering. For example, they can be used to model and analyze data, calculate probabilities, and solve optimization problems.
Yes, there are several rules and properties to keep in mind when working with sum binomials. Some of the most important ones include the distributive property, the commutative property, and the difference of squares formula. It is important to also remember to carefully simplify expressions and combine like terms when working with sum binomials.