Today I encountered the following identity during some work: letting $b_k$ be a sequence, $$ \sum_{l=1}^n(-1)^{l-1}\binom{n}{l}\sum_{k=0}^{l-1}\binom{l-1}{k}b_k = \sum_{k=0}^{n-1}(-1)^k b_k \,. $$ I've struggled for a while with properties of binomial coefficients and induction strategies, then I think I found a proof using derivatives (see below). Are there more conventional/straightforward proofs?
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Actually, there is a way by direct manipulation of the binomials. $$ \bbox[lightyellow] { \eqalign{ & \sum\limits_{l = 1}^n {\left( { - 1} \right)^{\,l - 1} \left( \matrix{ n \cr l \cr} \right)\sum\limits_{k = 0}^{l - 1} {\left( \matrix{ l - 1 \cr k \cr} \right)b_{\,k} } } = \quad \quad (1) \cr & = \sum\limits_{1\, \le \,l\,\left( { \le \,n} \right)} {\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,l - 1} \right)} {\left( { - 1} \right)^{\,l - 1} \left( \matrix{ n \cr l \cr} \right)\left( \matrix{ l - 1 \cr k \cr} \right)b_{\,k} } } = \quad \quad (2) \cr & = \sum\limits_{0\, \le \,l\,\left( { \le \,n - 1} \right)} {\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,l} \right)} {\left( { - 1} \right)^{\,l} \left( \matrix{ n \cr l + 1 \cr} \right)\left( \matrix{ l \cr k \cr} \right)b_{\,k} } } = \quad \quad (3) \cr & = \sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,n - 1} \right)} {\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,l} \right)} {\left( { - 1} \right)^{\,l} \left( \matrix{ n \cr l + 1 \cr} \right)\left( \matrix{ l \cr l - k \cr} \right)b_{\,k} } } = \quad \quad (4) \cr & = \sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,n - 1} \right)} {\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,l} \right)} {\left( { - 1} \right)^{\,l} \left( \matrix{ n \cr l + 1 \cr} \right)\left( { - 1} \right)^{\,l - k} \left( \matrix{ - k - 1 \cr l - k \cr} \right)b_{\,k} } } = \quad \quad (5) \cr & = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,l} \right)} {\left( {\sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,n - 1} \right)} {\left( \matrix{ n \cr l + 1 \cr} \right)\left( \matrix{ - k - 1 \cr l - k \cr} \right)} } \right)\left( { - 1} \right)^{\,k} b_{\,k} } = \quad \quad (6) \cr & = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,l} \right)} {\left( {\sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,n - 1} \right)} {\left( \matrix{ n \cr n - 1 - l \cr} \right)\left( \matrix{ - k - 1 \cr l - k \cr} \right)} } \right)\left( { - 1} \right)^{\,k} b_{\,k} } \quad \left| {\;0 \le n} \right.\quad = \quad \quad (7) \cr & = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n - 1} \right)} {\left( \matrix{ n - k - 1 \cr n - k - 1 \cr} \right)\left( { - 1} \right)^{\,k} b_{\,k} } \quad \left| {\;0 \le n} \right.\quad = \quad \quad (8) \cr & = \sum\limits_{0\, \le \,k\, \le \,n - 1} {\left( { - 1} \right)^{\,k} b_{\,k} } \quad \left| {\;1 \le n} \right. \quad \quad (9) \cr} }$$
where:
(1) original expression;
(2) we get rid (put in brackets) of the summation bounds
which are implicit in the binomial;
(3) we change $l$ with $l-1$;
(4) since $0 \le l$ we can apply symmetry and get rid
also of the lower bound on $l$;
(5) Upper Negation;
(6) exchange order of summation;
(7) symmetry on the first binomial, provided that $0 \le n$;
(8) bounds allow to apply convolution in $l$;
(9) the binomial provides the bounds on summation
Concerning the approach you propose, it looks "formally" pretty valid.
I am hesitant from a rigorous point of view when it comes to take the multiplicative inverse, but I am not a theorist.
However, yours parallels a slight different approach that I would take, based on the Right Shift Operator, which I know is valid also from a rigorous point of view.
$$ \bbox[lightyellow] { E\,f(x) = f(x + 1)\quad E\;b_{\,k} = b_{\,k + 1} \quad b_{\,k} = E^{\,k} \;b_{\,0} } $$ Then $$ \bbox[lightyellow] { \eqalign{ & LHS = \sum\limits_{l = 1}^n {\left( { - 1} \right)^{\,l - 1} \left( \matrix{ n \cr l \cr} \right)\sum\limits_{k = 0}^{l - 1} {\left( \matrix{ l - 1 \cr k \cr} \right)b_{\,k} } } = \sum\limits_{l = 1}^n {\left( { - 1} \right)^{\,l - 1} \left( \matrix{ n \cr l \cr} \right)\sum\limits_{k = 0}^{l - 1} {\left( \matrix{ l - 1 \cr k \cr} \right)E^{\,k} \;b_{\,0} } } = \cr & = b_{\,0} \sum\limits_{l = 1}^n {\left( { - 1} \right)^{\,l - 1} \left( \matrix{ n \cr l \cr} \right)\left( {I + E} \right)^{\,l - 1} } = - {{b_{\,0} } \over {\left( {I + E} \right)}}\sum\limits_{l = 1}^n {\left( { - 1} \right)^{\,l} \left( \matrix{ n \cr l \cr} \right)\left( {I + E} \right)^{\,l} } = \cr & = - {{b_{\,0} } \over {\left( {I + E} \right)}}\left( {\left( {I - \left( {I + E} \right)} \right)^{\,n} - I} \right) = - b_{\,0} {{\left( { - 1} \right)^{\,n} E^{\,n} - I} \over {\left( {I + E} \right)}} = b_{\,0} {{I - \left( { - 1} \right)^{\,n} E^{\,n} } \over {\left( {I + E} \right)}} \cr} } $$ and $$ \bbox[lightyellow] { RHS = \sum\limits_{0\, \le \,k\, \le \,n - 1} {\left( { - 1} \right)^{\,k} b_{\,k} } = b_{\,0} \sum\limits_{0\, \le \,k\, \le \,n - 1} {\left( { - 1} \right)^{\,k} E^{\,k} } = b_{\,0} {{I - \left( { - 1} \right)^{\,n} E^{\,n} } \over {I + E}} } $$
G Cab
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Your argument is very nice: the RSO coincides with the derivative when $b_k=\partial^k f$, so I guess this makes it rigorous. Thanks! – Brightsun Oct 26 '17 at 20:43
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Well, then instead of a "function" f(x), which would imply problems of convergence for some $b_k$ sequences, I guess you should talk of a "formal power series". – G Cab Oct 27 '17 at 14:31
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@G Cab What convergence issues are you referring to? I am just using the fact that every sequence is the sequence of derivatives of some smooth (but not necessarily analytic) function https://math.stackexchange.com/questions/10643/every-real-sequence-is-the-derivative-sequence-of-some-function – Brightsun Oct 27 '17 at 23:18
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@Brightsun: if, e.g., $b_k=(k!)^2$, then what "smooth function" will correspond to it ? – G Cab Oct 28 '17 at 15:06
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As far as I know the function is not unique, but nonetheless it exists: it's a result due to Borel (see the link in my previous comment). @G Cab – Brightsun Oct 29 '17 at 19:56
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@Brightsun: Uhm, I did not know that power series with a null convergence radius can still be named functions. But as I said I am not a theorist, so I am going to accept that. – G Cab Oct 29 '17 at 22:41
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@G Cab Indeed they cannot. The claim by Borel is that, given any sequence $b_k$, then one can find a function $f(x)$ such that $b_k=\partial^k f$, but it does not claim that $f(x)=\sum_k b_k x^k/k!$, in general. The latter is true only for those $b_k$ for which the series converges, as you correctly pointed out, and then the result is trivial. – Brightsun Oct 30 '17 at 14:32
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Thanks, now I got better understanding of "Borel's claim". However, it seems to me that it would be less troublesome to found your claim onto "formal power series", or .. is it not? – G Cab Oct 30 '17 at 16:37
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@G Cab A formal power series does not define a function. The Borel's theorem ensures you can find a function with the above property. – Brightsun Oct 30 '17 at 23:37
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@G Cab Also, $b_k=(k!)^2=\partial^k f$ is satisfied by $f(x)=\int_0^\infty e^{-t}/(1-xt)dt$, defined for $x<0$. – Brightsun Oct 31 '17 at 00:01
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We start from
$$\sum_{l=1}^n (-1)^{l-1} {n\choose l} \sum_{k=0}^{l-1} {l-1\choose k} b_k = \sum_{k=0}^{n-1} b_k \sum_{l=k+1}^n {n\choose l} (-1)^{l-1} {l-1\choose k}.$$
We get for the coefficient on $b_k$
$$\sum_{l=k+1}^n {n\choose l} (-1)^{l-1} {l-1\choose k} = \sum_{l=k}^{n-1} {n\choose l+1} (-1)^l {l\choose k} \\ = \sum_{l=k}^{n-1} {n\choose l+1} (-1)^l [z^k] (1+z)^l.$$
We may start $l$ at zero because $[z^k] (1+z)^l$ is zero when $0\le l\lt k$, getting
$$[z^k] \sum_{l=0}^{n-1} {n\choose l+1} (-1)^l (1+z)^l = - [z^k] \frac{1}{1+z} \sum_{l=0}^{n-1} {n\choose l+1} (-1)^{l+1} (1+z)^{l+1} \\ = - [z^k] \frac{1}{1+z} \left((1-(1+z))^n - 1\right) = [z^k] \frac{(-1)^{n+1} z^n}{1+z} + [z^k] \frac{1}{1+z}.$$
Note however that $n\gt k$ so there is no contribution from the first term and we are left with
$$[z^k] \frac{1}{1+z} = (-1)^k$$
as claimed.
Marko Riedel
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This method entails essentially the same cancellations as mine (especially the resummation which gives $(1-1-z)^n-1$) but from the perspective of the $[z^k]$ operator. I am only not really sure of why you got rid of $b_k$ all over the place...? – Brightsun Oct 26 '17 at 20:48
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Well on the first line the inner sum is the coefficient on $b_k$ so if we can prove it is $(-1)^k$ we have the claim independent of the values $b_k$ actually takes. – Marko Riedel Oct 26 '17 at 21:02
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For a given sequence $b_k$, by Borel's lemma (see here or here), there exists a function $f(x)$ such that $ b_k = f^{(k)}(0) \equiv \partial^kf. $ Thus we need only prove $$ \sum_{l=1}^n(-1)^{l-1}\binom{n}{l}\sum_{k=0}^{l-1}\binom{l-1}{k}\partial^k = \sum_{k=0}^{n-1}(-1)^k \partial^k\,. $$ The right-hand side can be evaluated formally using $1-x+x^2+\cdots+(-x)^{n-1}=(1-(-x)^n)/(1+x)$, and yields $$ \frac{1-(-\partial)^n}{1+\partial}\,. $$ The left-hand side can be manipulated using Newton's binomial formula, giving $$ \sum_{l=1}^n(-1)^{l-1}\binom{n}{l}(1+\partial)^{l-1}=\frac{1}{-1-\partial}\sum_{l=1}^n\binom{n}{l}(-1-\partial)^l=\frac{(1-1-\partial)^n-1}{-1-\partial}=\frac{1-(-\partial)^n}{1+\partial}\,. $$
Brightsun
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