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....
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.