In an expression like $$\sum_{k=1}^n (k^2-1)$$ I know (somehow, but how exactly is unclear) that $k$ is a real number, thus I can use all the applicable identities to rewrite it as $$\sum_{k=1}^n (k+1)(k-1)$$ But the reasoning should happen inside the expression, as $k$ is bound to the $\Sigma$, yet what I just did was outside of it. Casting this problem aside, I am not sure how I can use $k\in\mathbb R$. What indicates that it is a useable statement?


How to prove that

$$\sum_{k=1}^n (k^2-1)= \sum_{k=1}^n (k+1)(k-1)$$

  • $\begingroup$ If $a_k = b_k$ for all $k = 1,\ldots, n$, then $\sum\limits_{k=1}^n a_k - \sum\limits_{k=1}^n b_k = \sum\limits_{k=1}^n (a_k - b_k) =\sum\limits_{k=1}^n 0 = 0$ $\endgroup$ – Slugger May 8 '18 at 9:54
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    $\begingroup$ If you just compute the terms (in either of the two equivalent ways), you see that both sums are simply ways of writing $0+3+8+15+24+\dots$, so how can there possibly be any doubt that they are equal? $\endgroup$ – Hans Lundmark May 8 '18 at 10:01
  • $\begingroup$ @Slugger is this what you are thinking of, when you look at the equation? $\endgroup$ – Adam May 8 '18 at 10:10
  • $\begingroup$ Well if the result is not obvious and you want to convince yourself that both sums are equal, it would make sense to try and show that $\sum a_k - \sum b_k = 0$ to see that the sums are equal. I would normally agree with @HansLundmark and say that the result is just obvious since we are dealing with finite sums $\endgroup$ – Slugger May 8 '18 at 10:13
  • $\begingroup$ @HansLundmark on the contrary, it is trivial, and I want to understand why it is trivial, the same question could also be applied to e.g. $\{|x|:x\in\mathbb R\wedge x>0\}= \{x:x\in\mathbb R \wedge x>0\}$ $\endgroup$ – Adam May 8 '18 at 10:14

If you set out carefully the definition of what $S_n=\sum_{k=1}^{n} a_k$ means it may help.

I think it is defined recursively: $$ S_1:=a_1\ \text{and then}\ \ S_{n+1}:=S_n + a_{n+1}. $$

You can now prove that whenever $a_k=b_k$ for all $k$ then $\sum_{k=1}^{n} a_k=\sum_{k=1}^{n} b_k$. (Yours is the special case $a_k=(k^2-1)$, $b_k=(k-1)(k+1)$.)

Base case: $$\sum_{k=1}^{1} a_k= a_1 =b_1 =\sum_{k=1}^{1} b_k,$$ the equalities justified by definition of $\sum_{1}^{1}$, hypothesis, definition of $\sum_{1}^{1}$.

Inductive step: $$\sum_{k=1}^{n+1} a_k= \sum_{k=1}^{n} a_k +a_{n+1}= \sum_{k=1}^{n} b_k +a_{n+1}= \sum_{k=1}^{n} b_k +b_{n+1}= \sum_{k=1}^{n+1} b_k, $$ where the equalities are justified by definition of $\sum_{1}^{n+1}$, inductive hypothesis, hypothesis, definition of $\sum_{1}^{n+1}$.

In terms of the way you explain your difficulty, note that we have used the definition of the $\Sigma$ symbol to "unbind" the top term, and then operate on it as we please, before using the definition of $\Sigma$ to "rebind" it.

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  • $\begingroup$ Thanks @Mauro ALLEGRANZA. $\endgroup$ – ancientmathematician May 8 '18 at 11:59
  • $\begingroup$ is this really what happens in your mind? seems like there should be a middle ground between "trivial" and this answer, that would prove both the stated equality and $\{k^2-1:k\in \mathbb R\}=\{(k-1)(k+1): k\in \mathbb R\}$ $\endgroup$ – Adam May 8 '18 at 15:43
  • $\begingroup$ I'm not sure where your obsession with $k\in\mathbb{R}$ comes from. The notation $\sum_{k=1}^{n}$ means $k\in\mathbb{N}$. Showing the sets you mention are equal -- both are $[-1,\infty)$ -- is easy, but has nothing to do with the sums. $\endgroup$ – ancientmathematician May 8 '18 at 15:52
  • $\begingroup$ If you think about it a bit, $\sum_{k=1}^{n} a_k$ depends on the sequence $\{a_n\}$ and not the set of values $\{a_k| k=1,\dots,n\}$-- there can be repeats, and as we look at $S_1, S_2,\dots$ the order of the $a_n$ matters too. That's why a non-handwaving definition of $S_n$ is recursive; and that being so why any sensible proof will be by induction. $\endgroup$ – ancientmathematician May 8 '18 at 16:01
  • $\begingroup$ I guess I am loving me some reals. you wrote "$\sum_{k=1}^n$ means $k\in \mathbb N$ ", this is what I am really curious about, where does this implication come from (even though it is not a real implication)? $\endgroup$ – Adam May 8 '18 at 16:14

The thing that will help you is one of these properties of equality:
1)If $$x=y$$ then for any function $f$ $$f(x)=f(y)$$
2)If $$x=y$$ then for any prediacte $\phi$, $\phi(x)<=>\phi(y)$
With either of this you can prove what you want to.
With (1) you can take function $f$ such that for any $x$ $$f(x)=\sum_{k=1}^n x $$ So $$k^2-1=(k-1)(k+1)$$ Applying first property:$$\sum_{k=1}^n k^2-1 = \sum_{k=1}^n (k-1)(k+1)$$
With (2) you can take predicate $\phi$ such that $\phi(x)$ means $$\sum_{k=1}^n x = \sum_{k=1}^n (k-1)(k+1)$$ So $$(k-1)(k+1)=k^2-1$$ So if $\phi((k-1)(k+1))$ is true then $\phi(k^2-1)$ is true. But $\phi((k-1)(k+1))$ is true by reflexive property of equality because $\phi((k-1)(k+1))$ is $$ \sum_{k=1}^n (k-1)(k+1) = \sum_{k=1}^n (k-1)(k+1)$$ So we must have $\phi(k^2-1)$ also be true, so $$\sum_{k=1}^n k^2-1 = \sum_{k=1}^n (k-1)(k+1)$$ is true.

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