I have successfully proven $ \displaystyle \sum_{k=1}^n k ·(k!) = (n+1)! -1 $ with mathematical induction for all $n \in \mathbb{N}$. Now, how would someone prove this assertion without induction?
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2Technically you can't. But the induction may be hidden so that the proof of the induction step is so obvious a trivial it doesn't need stating, and so that the prover doesn't use the ritual incantation of "StatethepropositionasP(n)dothebasecasen=1thendotheinductivestepofassumingP(n)istruethenproveP(n+1)mustbetrue"... $\sum(kk!)=\sum_{k=0}^n((k+1)k!-k!)$ which "telescopes" to $(n+1)n! + \sum_{k=0}^{n-1}(k+1)! - \sum_{k=1}^{n}k! - 0!=(n+1)n!+(\sum_{k=1}^n(k!-k!))-1=(n+1)n!-1$. BUT!* that "telescoping" works for all $n$ is an assumption of induction. – fleablood Mar 21 '20 at 16:57
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@fleablood The only "hidden" induction in telescoping is that that the addition/subtraction of an arbitrary number of terms can be rearranged arbitrarily without affecting the outcome. If one is not equipped with this as a starting point, then there are a voluminous number of proofs that we "normally" don't consider inductive, would be inductive in part. – Mark Viola Mar 21 '20 at 17:40
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It presumes what the terms of the sums actually are. If presumes if you have $kk!$ as one term and that $kk! = (k+1)!-k!$ then the only sense you can make of that is if you assume that the next term is $(k+1)(k+1)!$ and the previous term is $(k-1)(k-1)!$ and you absolutely can NOT assume that without induction.. And I would say we do consider those voluminous number of proofs to be inductive. Very very very few proofs are not inductive. – fleablood Mar 21 '20 at 23:49
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If you are not supposed to deliver a "formal proof" in which every simple fact is rigorously proven, the below method does not involve induction. Apparently, for some mathematicians, mathematics only exists if everything is formalized. My philosophy is different. – Peter Mar 22 '20 at 09:11
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Hint : $$k\cdot k!=(k+1)\cdot k!-k!=(k+1)!-k!$$ Almost all terms will cancel out.
Peter
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1But doesn't the proof that $\sum_{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1$ use induction? (Well, I suppose you could go through coefficients to verify what survives, but then you need to prove by induction that you can sum $n+1$ terms in any order you like.) – J.G. Mar 21 '20 at 16:22
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No, we have $$a_{n+1}-a_n+a_n-a_{n-1}+a_{n-1}-a_{n-2}+...a_3-a_2+a_2-a_1=a_{n+1}-a_1$$ I do not think that we need to prove the associativity of the addition. – Peter Mar 21 '20 at 16:25
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Consider the $n=3$ case, $(a_3-a_2)+(a_2-a-1)=a_3+(-a_2+a_2)-a_1$. Now consider successively larger $n$. Are you sure you can make the general case rigorous without induction? – J.G. Mar 21 '20 at 16:33
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@J.G. That we can shift the paranthesis is a well known property. Anyway, I am sure that the trick I mentioned is the desired answer how to proof the claim without induction, whether if this is actually somewhere hidden is rather a matter of taste. To my opinion, there is no induction. – Peter Mar 21 '20 at 16:37
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3Yeah.... "almost all terms will cancel out" is an inductive statement. To my mind, this is a proof by induction. Any assumption that "$...$" can be extended indefinitely is relying upon the induction hypothesis. – fleablood Mar 21 '20 at 16:38
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You can downvote the answer, if you want and really insist that this is an induction. But I think, you agree that the trick is what is supposed to be noticed by the student. – Peter Mar 21 '20 at 16:41
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@J.G. $$\begin{align}\sum_{k=1}^n (a_{k+1}-a_k)&=\sum_{k=1}^n a_{k+1}-\sum_{k=1}^n a_k\\&=\sum_{k=2}^{n+1}a_k-\sum_{k=1}^n a_k\\&=\left(\sum_{k=1}^na_k +a_{n+1}-a_1\right)-\sum_{k=1}^n a_k=a_{n+1}-a_1\end{align}$$Was induction required here? – Mark Viola Mar 21 '20 at 17:17
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@MarkViola I'd say yes, because how do you know the rearrangement in$$\sum_{k=1}^n(b_k-c_k)=\sum_{k=1}^nb_k-\sum_{k=1}^nc_k$$works for all $n$? The rearrangement of $\sum_{k=2}^{n+1}a_k$ seems to need induction as well. – J.G. Mar 21 '20 at 17:19
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1@J.G. I assume that we have already proved that rearrangement works for all $n$. If we need to begin from first principles, then many proofs are in part inductive. – Mark Viola Mar 21 '20 at 17:23
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@Peter It is a matter of "starting points." That is to ask, "where do we begin?" If we don't take for granted the rearrangement property of addition/subtraction, then many proofs that we would not "normally" classify as inductive, become inductive in part. – Mark Viola Mar 21 '20 at 17:27
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@MarkViola Moreover, the exercise is clearly to prove it (in one part) without induction. This clearly indicates , that the author of the exercise does NOT consider this trick to be induction. Otherwise the exercise would not make sense. – Peter Mar 21 '20 at 17:29
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@MarkViola Your & Peter's pragmatism does seem likely to be necessary. After all, one could even argue "pedantic" induction is needed to make factorials well-defined. Boy, PA- is limited. – J.G. Mar 21 '20 at 17:58
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@J.G. Indeed. As I mentioned in previous comments, there would be a voluminous set of proofs that we "normally" do not view as inductive in nature, that would be inductive in part (at least). – Mark Viola Mar 21 '20 at 18:01
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@MarkViola It raises the question of whether we should define a PA-/PA intermediate with just enough induction for telescoping. Of course, even if telescoping isn't generally called indicative, its results are very common choices for exercises in induction, to the point there's a great question asking for other uses of induction. – J.G. Mar 21 '20 at 18:09
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@J.G. This exercise is not about the peano axioms. We can assume the simple facts and need not reinvent the wheel just to can claim that induction cannot be avoided. If this were an exercise about formal logic, this might be different. – Peter Mar 22 '20 at 09:06
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@Peter How do you define weak induction on the non-negative integers, other than the way PA does? – J.G. Mar 22 '20 at 09:18