When do regularization methods for divergent series disagree? Sometimes, it is possible to take a divergent series (in the sense of its sequence of partial sums failing to converge) and "regularize" it using one of a variety of methods to assign it a meaningful finite value. For example, by observing that $\zeta(-1) = -1/12$ via analytic continuation, we can claim in some sense that
$$ 1 + 2 + 3 + 4 + \cdots = \sum_{k=1}^\infty k = -\frac{1}{12}. $$
Remarkably, different methods of regularization applied to this divergent series all give the same result, suggesting that the answer $-1/12$ is no "accidental" consequence of the method of regularization.
My question is, what are some natural examples of divergent series for which this fails to occur, i.e., for which two regularization methods give different finite answers?
 A: This doesn't really answer your question, but it goes in to a bit of depth on how regularization works. (Does "a bit of depth" make sense?)
So we want to sum a divergent series. Let's assume there's nothing wrong with adding term-by-term, or by putting a zero in front of a series. Here are a few example summations:

To sum $1-1+1-1+\dotsb$:
  $\,\;s=1-1+1-1+\dotsb$
  $\,\;s=0+1-1+1-\dotsb$ (putting a zero in front)
  $2s=1+0+0+0+\dotsb$ (summing term-by-term)
  Thus, $1-1+1-1+\dotsb=\dfrac12$.

To sum $1-2+3-4+\dotsb$:
  $\,\;s=1-2+3-4+5-\dotsb$
  $\,\;s=0+1-2+3-4+\dotsb$
  $\,\;s=0+1-2+3-4+\dotsb$
  $\,\;s=0+0+1-2+3-\dotsb$
  $4s=1+0+0+0+0+\dotsb$ (summing term-by-term)
  Thus, $1-2+3-4+5-\dotsb=\dfrac14$. (By the way, this agrees with $\eta(-1)=\dfrac14$, where $\eta$ is the alternating zeta function.)

As my first comment shows, it is impossible to do this with your sequence:

$\,\,\;s=1+2+3+4+5+\dotsb$
  $-s=0-1-2-3-4-\dotsb$ (multiplying by -1 and adding zero)
  $-s=0-1-2-3-4-\dotsb$
  $\,\,\;s=0+0+1+2+3+\dotsb$
  $\,\,\,0=1+0+0+0+0+\dotsb$
  A contradiction.

So, if you want to sum that series, you have to assume that either putting a zero in front of a series can change a value, or that summing term-by-term is problematic.

By the way, I should mention: There are several ways to assign a value to a divergent series. One is Cesàro summation ("che-SAH-ro"). Let's take $1-1+1-\dotsb$ as an example. Its partial sums are $1,0,1,0,$ etc. To find the Cesàro sum, just take the average. (In notation: If $s_n$ is the nth partial sum, then the Cesàro sum is $\lim_{N\to\infty}\frac{s_0+\dotsb+s_N}{N}$.) This easily gives us $1-1+\dotsb=\dfrac12$.
This doesn't help us sum all series, however. The Cesàro sum of $1-2+3-\dotsb$ is undefined. Another way, then, is to do the Cesàro sum twice: Let $t_n$ be $\frac{s_0+\dotsb+s_N}{N}$. Then, we can say that the sum is $\lim_{N\to\infty}\frac{t_0+\dotsb+t_N}{N}$. (I don't know what the name of this is.) This gives us the result $1-2+3-\dotsb=\dfrac14$. (I won't show you the calculations.)
Another way is to make a power series: If we want to sum $a_0+a_1+\dotsb$, see if the function $a(x)=a_0+a_1x+a_2x^2+\dotsb$ converges for any $x$. Then, set $x=1$. For example: Since $1-x+x^2-x^3+\dotsb=\dfrac1{x+1}$, plugging in $x=1$ gives us $1-1+1-\dotsb=\dfrac12$. A similar calculation shows us that $1-2+3-\dotsb=\dfrac14$. However, we run into problems with $1+2+3+\dotsb$. $1+2x+3x^2+\dotsb=\dfrac x{(x-1)^2}$; plugging in $x=1$ doesn't give us a value. Oh, no.
It should be said that none of those methods contradict each other; if two methods can sum a series, then they'll give the same value. Also, if a series is convergent, all of those methods give it the correct sum.
