Exercising divergent summations: $\lim 1-2+4-6+9-12+16-20+\ldots-\ldots$

I’m trying to make sense of some (assumed to be) simple exercises in divergent summation. One example I cannot resolve.
First I assume the sequence of binomialcoefficients $ \{ b_k = \binom k2 \}_{k=2\to\infty}=\{1,3,6,10,15,…\}$
Then to assign a meaningful value to the alternating sum $ S= 1-3+6+10…. $ I compute the Abel-sum

$ \qquad S = \lim_{x \to 1} 1-3x+6x^2-10x^3+…-… = {1 \over (1+x)^3 } = 1/8 $ (Abel)

But I want to proceed one more step. The sequence of partial sums may be denoted as

$ \{ c_k \}_{k=0\to\infty}=\{1,-2,4,-6,9,-12,16,-20,…\} $

Q – my question is: what is the sum $ T = 1-2+4-6+9… $ ?


I tried two approaches, but I’m lost.
The generating function is simply $ 1/(1+x)^3/(1-x) $ but here I cannot let x approach 1, so the simple application of the Abel-sum is impossible. Also the Euler-sum seems to not to converge; instead I get increasing partial sums like k/16 for the k’th partial sum .

On the other hand, I observe, that the coefficients are near the squares, so if I consider the alternating $\zeta$ (usually called Dirichlet-$\eta$)

$ \qquad \qquad \begin{eqnarray} X &=& 1 – 2.25 + 4 – 6.25 + 9 – 12.25 + … \\ &=&( 4 – 9 + 16 – 25 + … – …)/4 \\ &=& (1-\eta(-2))/4 \\ &=& (1-0)/4 = {1 \over 4} \end{eqnarray}$

then in T I had just each second coefficient $1/4$ above that of the $\eta$-series in X and possibly could go along something like

$ \qquad \qquad \begin{eqnarray} T &=& 1 – (2.25-0.25) + 4 – (6.25-0.25) + … – … \\
&=& X + 0.25*\zeta(0) \\
&=& X-1/8 = {1 \over 8} \end{eqnarray}$

But surely this is only an outline how I could come nearer to a solution. How could I actually proceed here?

[update]: One more idea was to make use of reordering summation. The coefficients $c_k$ can be seen as rowsums of the following matrix:

$\qquad \small
\begin{array} {rrrrr}
1 & . & . & . & . & . &\ldots\\
-3 & 1 & . & . & . & . \\
6 & -3 & 1 & . & . & . \\
-10 & 6 & -3 & 1 & . & . \\
15 & -10 & 6 & -3 & 1 & . \\
-21 & 15 & -10 & 6 & -3 & 1 & \ldots \\
… & … \\
\end{array}
$

so that all columns evaluate to $1/8$ due to the Abel-summation. If I add all that columnsums, I should have to write $\zeta(0)*1/8 $ and evaluate $T=-1/16$ (Q&D) now. But this is all fumbling, because I not even reflect the infinite application of downshifting by rows when evaluating the columns…

[update2]: Hmm. I played with the reciprocals of the series. I just did the “paper&pen” divisions and got:
$ \small \qquad \qquad 1 \qquad : 1-3x+6x^2-10x^3+15x^4-21x^5+…-…=1+3x+3x^2+1x^3=(1+x)^3=|_{x=1}8 $
and
$ \small \qquad \qquad 1 \qquad : 1-2x+4x^2-6x^3+9x^4-12x^5+…-…=1+2x-2x^3-1x^4=(1+x)^3(1-x)=|_{x=1}0 $
So this is also immediately what Lubos pointed out. Good for my intuition, I hope this model is not too much misleading in other obvious cases….

Solutions Collecting From Web of "Exercising divergent summations: $\lim 1-2+4-6+9-12+16-20+\ldots-\ldots$"

Dear Gottfried, as you correctly observe, the sum is the Taylor expansion of
$$ \frac{1}{(1+x)^3 (1-x)} $$
for $x=1$. This function has a pole at $x=1$, so the result is a genuine divergence, the standard number “infinity” (without a specification of the phase) that is inverse to zero. The fact that some seemingly divergent sums have finite values doesn’t mean that all of them have finite values.

Your second method is illegitimate because it clumps the neighboring values of $n$ – the exponents in the powers of $x$ – which means that the justification isn’t robust under any infinitesimal deformations of the parameters. Note that it wasn’t really legitimate that you wrote $1+1+1+\dots = \zeta(0)$. In fact, $\zeta(0)-n$ for any integer $n$ – and in fact, not only integer – would be equally (un)justified. In fact, none of them gives the right result.

It’s a misconception that $1+1+1+\dots = -1/2$ “always” holds. It’s only true if the terms $1$ are associated with values of $n$ that go over positive integers. But if they go over non-negative integers, the result would be $+1/2$, and so on.

I have an answer, applying a principle of which I do not yet know how far this is applicable in general or whether at all. It assumes, that as far as a given series can be expressed as linear combination of (unshifted) formal zeta- and (Dirichlet)-eta series then the evaluation of the series can be taken by evaluation of the same combination of zeta and eta-values, even if divergent (possibly with the exception of the zeta at 1).

So this would give the following regularized result:

Denote the original series by $T$, then let
$$ f(s) = {1\over 1^s }-{2\over 2^s}+{4\over 3^s}-{6\over 4^s}+{9\over 5^s}- \cdots + \cdots \tag 1$$ of course attempting to justify $ T = \lim_{s \to 0} f(s)$.

This is convergent for $s \gt 3$ . For this cases we can decompose
$$ \begin{array}{rcrll} 8 f(s) &=& & {8\over 1^s }-{16\over 2^s}+{32\over 3^s}-{48\over 4^s}+{72\over 5^s}- \cdots + \cdots \\
&=& 1(&{1\over 1^s }+{1\over 2^s}+{1\over 3^s}+{1\over 4^s}+{1\over 5^s}+ \cdots + \cdots &)\\
& & + 1(&{1\over 1^s }-{1\over 2^s}+{1\over 3^s}-{1\over 4^s}+{1\over 5^s}- \cdots + \cdots&) \\
& & + 4(&{1\over 1^s }-{2\over 2^s}+{3\over 3^s}-{4\over 4^s}+{5\over 5^s}- \cdots + \cdots & ) \\
& & + 2(&{1\over 1^s }-{4\over 2^s}+{9\over 3^s}-{16\over 4^s}+{25\over 5^s}- \cdots + \cdots & ) \\
&\underset{s \gt 3}=& &1 \zeta(s)+1\eta(s)+4\eta(s-1)+2\eta(s-2)
\end{array} \\ \phantom{dummy } \\
f(s) \underset{s \gt 3} = {\zeta(s)+\eta(s)\over 8} + {\eta(s-1)\over 2} + {\eta(s-2)\over 4} \qquad \qquad \qquad \tag 2
$$
and from this, assuming it is regularizable for setting $s=0$
$$ T = f(0) \underset{\mathcal Z}{=} {\zeta(0)+\eta(0)\over 8} + {\eta(-1)\over 2} + {\eta(-2)\over 4} = 0 + {1\over4}\cdot{1\over 2} + 0 = {1\over 8}
\tag 3$$
$\qquad \qquad $ where “$\mathcal Z$” means zeta-regularization

This is surely a nice consideration and general scheme, however I don’t know from which general theorems for divergent series I can extract this as an explicite recipe (and for what, possibly limited, cases).

The elementary Ramanujan sum of $1-2+4-6+9-\cdots$ is 1/4 and the series belongs to the class $R=2$.

In general, for $f(x)=b(x)/(1-x)$ and $b(x)$ Abel summable, the sum of $f(1)$
is $-b'(1) + b(1)/2$, with $b'(x)=xdb(x)/dx$.

Edit 1. Verification.
$$
\frac{1}{(1+x)^3(1-x)}=\frac{x}{8(1-x)}+\frac{8+7x+4x^2+x^3}{8(1+x)^3}~.
$$
The first term on the right-hand side corresponds to
$$
\frac{1}{8}(1+1+1+\cdots)=-\frac{1}{16}
$$
and the second term is Abel summable to $5/16$ (the value at $x=1$).
Linearity implies that the sum is $-1/16 + 5/16 = 1/4$.

Edit 2. The shifted series $0+1-2+4-\cdots=1/8$.

The shifted series corresponds to $x$ times the previous series, thus
$$
\frac{x}{(1+x)^3(1-x)}=\frac{x}{8(1-x)}-\frac{x}{8}+\frac{8x+7x^2+4x^3+x^4}{8(1+x)^3}~.
$$
As before, linearity implies that the sum is $-1/16-1/8+5/16=1/8$.

Let me try to give one more method for the computation of this divergent series, which resorts to $\zeta$-function regularization only in the very last step.

First, let us recall that the finite sums
$$
\mathcal T_p(N) \equiv \sum_{m=1}^N (-1)^{m-1} m^p = 1-2^p+3^p+\ldots \pm N^p
$$
can be computed as the derivatives at the origin of the generating function
$$
G_N(z) =\sum_{p=0}^\infty \mathcal T_p(N) \frac{z^p}{p!}=\sum_{m=1}^N(-1)^{m-1}\sum_{p=0}^\infty \frac{(mz)^p}{p!}=
\sum_{m=1}^N(-1)^{m-1}e^{mz}=\frac{1+(-1)^{N+1} e^{Nz}}{1+e^{-z}},
$$
In particular
$$\begin{aligned}
\mathcal T_1(N)=&\ 1-2+3-4+\ldots\pm N=\frac{(-1)^{N+1}(2N+1)-1}{4}, \\
\mathcal T_2(N)=&\ 1-4+9-16+\ldots\pm N^2=\frac{(-1)^{N+1}}{2}N(N+1).
\end{aligned}$$
Furthermore, the formal limit $N\to\infty$ gives the Euler sum of the corresponding divergent series:
$$
s_p=\sum_{n=1}^\infty (-1)^{m-1}m^p=1-2^p+3^p+\ldots
$$
are obtanied from the generating function
$$
G(z)=\sum_{p=0}^\infty s_p \frac{z^p}{p!}=\frac{1}{1+e^{-z}},
$$
which gives
$$\begin{aligned}
s_0=&\ 1-1+1-1+\ldots=\frac{1}{2}\ ,\\
s_1=&\ 1-2+3-4+\ldots=\frac{1}{4}\ ,\\
s_2=&\ 1-4+9-16+\ldots=0.
\end{aligned}
$$
After these preparations we get,
$$
S=\sum_{n=2}^\infty (-1)^n\frac{n(n-1)}{2}=\frac{1}{2}(s_1-s_2)=\frac{1}{8},
$$
and
$$
T=\sum_{n=2}^\infty \sum_{l=2}^n (-1)^l \frac{l(l-1)}{2}=\frac{1}{2}\sum_{n=1}^\infty\left( \mathcal T_1(n)-\mathcal T_2(n)\right)=\frac{1}{8}\sum_{n=1}^\infty((-1)^{n-1}-1)=\frac{1}{8}.
$$
In the last step we have used both that $s_0=1/2$ and $\zeta(0)=-1/2$.