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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 $0!=1$. – Quinn Culver Mar 27 '12 at 2:03
up vote 5 down vote accepted

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. – Gerry Myerson Mar 27 '12 at 2:03
@GerryMyerson: Did you mean to comment on my answer or on the original question? – MJD Mar 27 '12 at 2:06
@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. – MJD Mar 27 '12 at 2:14
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? :-P – Neil Traft Mar 27 '12 at 2:19
@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. – MJD Mar 27 '12 at 2:26

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