Let’s start with something easier. Suppose $$T_n\equiv\sum^n_{i=0}{\left(-1\right)^i}=1-1+1-1+\cdots+\left(-1\right)^n$$ Obviously, $T_n=1$ if $n$ is odd, and $T_n=0$ if $n$ is even. What is $T_\infty$?

Consider the polynomial $$t\left(x\right)\equiv\sum^\infty_{i=0}{\left(-1\right)^ix^i}=1-x+x^2-x^3+\cdots$$ so that $t\left(1\right)=1-1+1-1+\cdots=T_\infty$. However, $$t\left(x\right)=1-x\left(1-x+x^2-x^3+\cdots\right)=1-xt\left(x\right)$$ which implies $t\left(x\right)=1/\left(1+x\right)$. Consequently, $$T_\infty=t\left(1\right)=\frac{1}{2}$$ Now what? Let us move on to something trickier. Consider $$ U_n\equiv\sum^n_{i=0}{\left(-1\right)^ii}=0-1+2-3+\cdots+\left(-1\right)^nn $$ Following our previous strategy, we define $$u_0\left(x\right)\equiv\sum^\infty_{i=0}{\left(-1\right)^iix^i}=0-1x+2x^2-3x^3+\cdots$$ so that $u_0\left(1\right)=0-1+2-3+\cdots=U_\infty$. Let us also define $$u_1\left(x\right)\equiv\sum^\infty_{i=1}{\left(-1\right)^{i-1}\left(i-1\right)x^i}=0x-1x^2+2x^3-3x^4+\cdots$$ so that $u_1\left(1\right)=0-1+2-3+\cdots=U_\infty$. This means that $2U_\infty=u_0\left(1\right)+u_0\left(1\right)$. But $$ u_0\left(x\right)+u_1\left(x\right)=0-1x+1x^2-1x^3+1x^4\cdots=t\left(x\right)-1 $$ so $u_0\left(1\right)+u_0\left(1\right)=t\left(1\right)-1=-1/2$ and therefore $$ U_\infty=\frac{1}{2}\left[u_0\left(1\right)+u_0\left(1\right)\right]=-\frac{1}{4} $$ Okay. That’s kind of weird, but where does that get us?

Once more, we define $$ s\left(x\right)\equiv\sum^\infty_{i=0}{ix^i}=0+1x+2x^2+3x^3+\cdots $$ so that $s\left(1\right)=1+2+3+\cdots=S_\infty$. Noting $$ s\left(x\right)+u_0\left(x\right)=0+0x+4x^2+0x^3+8x^4+\cdots=4s\left(x^2\right) $$ and specifically that $s\left(1\right)+u_0\left(1\right)=4s\left(1\right)$ we find $$ S_\infty=s\left(1\right)=\frac{1}{3}u_0\left(1\right)=\frac{1}{3}U_\infty=-\frac{1}{12} $$ Did we not already establish that $S_\infty$ is

*larger*than any natural number? Apparently, $-1/12$ is a very, very large number.

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