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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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    @GerryMyerson : "tl;dr" means I win, right? :-D2012-03-30