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$\newcommand{\kinv}{(\frac{1}{k})}$ Take the equation $S_k = \frac{k-1}{k!}(\kinv^0+\kinv^1+\kinv^2+\cdots)$. Evaluate it for $k=1$, and we can see that $k-1=1-1=0$, so the whole thing must equal $0$.

Good. Now solve the geometric series and we find that $S_k = \frac{k-1}{k!}(\frac{k}{k-1})$. But if we evaluate this at $k=1$, then $S_k$ is undefined because we have $0$ in the denominator!

Okay, so what to do? Let's simplify to this: $S_k = \frac{k}{k!}$. So now we can evaluate it once again, and it is not undefined, but rather $1/1!=1$. Wait, what? Our previous, equivalent formula gave us $0$, not $1$!

Okay, what if we cancel the k's? Then we'd have $S_k = \frac{k}{k(k-1)(k-2)\cdots} = \frac{1}{(k-1)!}$. And here, once again, we have a $0$ in the denominator if we try to evaluate, and so the solution is once again undefined. :-(

Can someone explain why these equations don't always evaluate the same, even though they are valid and equivalent?

Correction: As others have pointed out, the 4th iteration of the formula actually evaluates to $1$.

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    When you say "Okay, what if we cancel the $k$'s...once again we have a $0$ in the denominator", you seem to be mistakingly using that $0!=0$; actually $0!=1$.2012-03-27

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They don't evaluate differently. If you plug in $k=1$ into the first expression, the $k-1\over k!$ becomes 0, as you said, but that does not mean you can conclude that the product is 0, because the $\left(\left({1\over k}\right)^0 + \left({1\over k}\right)^1 + \left({1\over k}\right)^2 + \cdots \right)$ becomes $1+1+1\cdots = \infty$. The whole expression is therefore equal to $0\cdot\infty$, which is, guess what, undefined, because of exactly this sort of situation.

This kind of situation comes up all over the place. For example, consider the infinite summation: $$\sum{1-1+1-1+\cdots}$$

Grouping the terms one way, they all cancel: $$(1-1)+(1-1)+\cdots = 0$$

But grouping them differently, the sum seems to be 1: $$1-(1-1)-(1-1)-\cdots=1$$

The conclusion is that the series can't be handled sensibly or consistently, and we say that this "summation" is indeterminate. A large part of the branch of mathematics known as analysis is devoted to determining when such questions make sense and can be handled consistently.

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    Moreover, the convention is that 0-factorial is 1, so your last effort doesn't have zero in the denominator.2012-03-27
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    @GerryMyerson: Did you mean to comment on my answer or on the original question?2012-03-27
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    @NeilTraft: You might also consider the expression $k^2-3k+2\over k-2$. If you put in $k=2$, you get $0/0$, which is undefined. But if you factor the numerator as $(k-2)(k-1)\over k-2$ and then cancel the common factor of $k-2$, you get just $k-1$, which is defined everywhere. So here we seem to have two examples where one is defined at $k=2$ and the other isn't. The answer in this case is that the cancellation step is legal only when the common factor is nonzero; that is, *only* when $k\not=2$, again for precisely this reason.2012-03-27
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    Thinking back, I now realize that I've tripped over this problem again and again when evaluating algebraic expressions. Are there some basic rules to keep in mind to stop me from making illegal steps?? Or would I have to study this whole branch of mathematics to know the difference? :-P2012-03-27
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    @NeilTraft: The big one is "you may not divide by zero". If you have zero in the denominator, the whole thing is ruined. So if for example you have $${k-1\over k!}{k\over k-1}$$ and you want to simplify to $k\over k!$ you have to add the caveat "…unless $k=1$, in which case the expression is undefined." More generally, any undefined subexpression (such as $\log x$ where $x≤0$ for instance) will spoil the whole computation. Perhaps you should post that as a separate question.2012-03-27
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    "your" referred to OP, not to you, Mark.2012-03-27
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    $0!=1$ is not merely a convention; it's a fact.2012-03-27
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    @Michael, http://en.wikipedia.org/wiki/Factorial says "The value of 0! is 1, according to the **convention** (emphasis added) for an empty product." And we all know Wikipedia is never wrong.2012-03-28
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    @GerryMyerson : This is something that mathematicians generally don't do very well. Maybe something needs to be done about that. That we write integrals as $\displaystyle\int_a^b f(x)\;dx$ rather than as $\displaystyle\overset{x=b}{\underset{x=a}{\Omega}} \{dx\;f(x)\}$ is an example of a convention. I think it's demonstrable that taking $0!$ to be anything but $1$ would at best complicate things pointlessly, thereby falsely making something simple appear to have complexities that aren't really there.2012-03-29
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    @Michael, to take 0-factorial to be anything other than 1 would certainly be a terrible idea; the alternative to taking it to be 1 is refusing to countenance it at all. One has the option of refusing to recognize empty products, or defining them to be 1. The convention is to define them to be 1.2012-03-29
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    @GerryMyerson : It is demonstrable that leaving $0!$ undefined makes things appear more complicated than they really are. That's not just a convention.2012-03-29
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    @Michael, not sure I follow you. Writing the Riemann zeta function as $\chi^{1,(2),\infty}_{\aleph,\xi}(s)$ makes things appear more complicated than they are, but writing it as $\zeta(s)$ is just a convention.2012-03-30
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    @GerryMyerson : A Senate committee consists of 6 Republicans and 8 Democrats. A subcommittee is chosen at random, all choices being equally probable. What is the probability that a majority are Democrats? Answer: one has [3 D & 2 R] or [4 D & 1 R] or [5 D & 0 R]. Find several probabilities and do the multiplications and additions. BUT one could have said [3 D & 2 R] or [4 D & 1 R] or [5 D]. In the last term in the numerator, one multiplies the number of way to pick 5 Democrats out of 8 by the way to choose 0 Republicans out of 6. Or alternatively, one simply uses the number of.....2012-03-30
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    .....ways to choose 5 Democrats out of 8. Now: If the number of ways to choose 0 Republicans out of 6 is not 1, then we have a contradiction. But if we leave the number of ways to choose 0 Republicans out of 6 undefined, then we have a complication. You're talking about a merely notational complication. But the complication we're talking about here is actually in effect being asserted to be a complexity in the problem itself. But it isn't really there. It is a demonstrable fact---not just a convention---that only using $\binom 6 0 = 1$ allows you to see the simplicity of the problem.2012-03-30
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    @MichaelHardy, tl;dr.2012-03-30
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    @GerryMyerson : "tl;dr" means I win, right? :-D2012-03-30