# 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?

• Doubling, we get $2+4+6+\dotsb=-\frac16$. Putting a zero in front of the original sequence, we have $0+1+2+\dotsb=-\frac1{12}$. Subtracting term-wise, we have $2+3+4+\dotsb=-\frac1{12}$. Adding one, we have $1+2+3+4+\dotsb=\frac{11}{12}$, which contradicts what we had earlier. – Akiva Weinberger Dec 30 '14 at 13:37
• Hmm, I think there was the same or at least very similar question here (or in mathoverflow) by Max Muller in 2012 or 2013, but don't have it at the moment – Gottfried Helms Dec 31 '14 at 0:36
• The example of @columbus make it easily clear, that regularization does not mean simply to play with the symbols - but to restrict oneself to methods, where the results become/stay consistent with the algebra and arithmetic. One of the most important restriction is (in my opinion) the indexing of the terms of the series. For instance:by consecutive powers of x, then finding whether there is a continuous interval of x, where this is convergent and the result can be determined. After possibly meaningful algebraic transformations then let x go to 1 - as far as no singularity occurs and evaluate. – Gottfried Helms Dec 31 '14 at 0:58
• @GottfriedHelms It seems like you're saying we should evaluate $\lim_{x\to1^-}1+2x+3x^2+\dotsb$. The problem, though, is that that gives us infinity. It should be mentioned, though, that the alternating series gives us $1-2+3-4+\dotsb=\frac14$, which agrees with $\eta(-1)=\frac14$ ($\eta$ being the alternating zeta function). – Akiva Weinberger Dec 31 '14 at 1:10
• @columbus: yes, if we would evaluate $\lim 1+2x+3x^2+...$ approaching $x \to 1^-$ we'd get an infinity. Thus we say: with that method we cannot regularize such series. But if we use $1^s + 2^s + 3^s + ...$ there is an (even infinite) continuous interval for $s$ where this is convergent. After that we can do some algebraic manipulation (converting it to an alternating series of this form) or apply Helmut Hasse's binomial composition and find a meaningful finite value. As far as I know we have the only disagreement in the methods that sometimes a concurring method gives an infinity – Gottfried Helms Dec 31 '14 at 9:05

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)
$-s=0-1-2-3-4-\dotsb$
$\,\,\;s=0+0+1+2+3+\dotsb$
$\,\,\,0=1+0+0+0+0+\dotsb$

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.

• PS: $1-1+1-1+\dotsb=\frac12$. However, adding in a lot of zeroes can change the sum: You can show that $1+0-1+1+0-1+\dotsb=\frac23$ and that $1-1+0+1-1+0+\dotsb=\frac13$. – Akiva Weinberger Dec 31 '14 at 3:40
• The Cesaro sum is the limit of the partial sums of the partial arithmetic means. That may be the term you were looking for. – Ian Dec 31 '14 at 3:42
• Wikipedia: "In 1890, Ernesto Cesàro stated a broader family of summation methods which have since been called (C, α) for non-negative integers α. The (C, 0) method is just ordinary summation, and (C, 1) is Cesàro summation as described above." The second method—the one with the $t_n$'s—is called (C, 2). Iterating again gives us (C, 3). – Akiva Weinberger Dec 31 '14 at 3:46
• Sorry I got the order reversed. – Ian Dec 31 '14 at 3:54
• You can also prove that $\eta(-2)=0$. Let $S=\eta(-2)=1-4+9-\dotsb$. We know that $S=1-4+9-\dotsb$. $3S=0+3-12+27-\dotsb$. $3S=0+0+3-12+27-\dotsb$. $S=0+0+0+1-4+9-\dotsb$. Adding term-by-term, we — miraculously — get $8S=1-1+0+0+0+\dotsb$! Thus, $8S=0$ and $S=0$. QED. – Akiva Weinberger Dec 31 '14 at 4:19