<s>[INST] 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....
 [/INST] 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 $-1/16-1/8+5/16=1/8$.
</s>