"Incorrect" derivation for sum of infinite natural numbers

Question

I am trying to derive sum of infinite natural numbers for which the established answer is $-1/12$ but I am getting $-1/6$ as my answer and am unable to figure out what exactly am I doing wrong. Probably something very silly or something very fundamental is wrong in my derivation. The way I proceed is:$$S = 1+2+3+4+5+6...$$ $$=> S = 1+3+5+7+9+... + 2+4+6+8+10+...$$ $$=> S = 1+3+5+7+9+... +2(1+2+3+4+5+...)$$ $$=> S = 1+3+5+7+9... + 2S$$ $$=>-S = 1+3+5+7+9...$$ shifting RHS a position to the right and adding to itself: $$-2S = 1+4+8+12+16+...$$ $$=>-2S = 1+4(1+2+3+4+5...)$$ $$-2S = 1+ 4S$$ which results in $S=-1/6$. It would be nice if someone can tell me which step of derivation is wrong and why is it wrong. Thanks for your help.

Answer 1

Let $$A = S-S = 1+(2-1)+(3-2)+(4-3)+\ldots = 1+1+1+1+\ldots$$ But $$n+A = (\underbrace{1+\ldots+1}_n)+1+1+1+1+\ldots= 1+1+1+1+\ldots= A$$ hence assigning a value to such divergent series is inconsistent and from $A= A+1$ you can obtain the value you'd like :

$$S = S+c(A-A) = S+c(A+1)-cA =S+c + c(A-A)= S+c$$

Now if you consider instead some divergent series summation method, for example $$T = \lim_{z \to 1^-} \sum_{k=1}^\infty (-1)^k k z^k$$ it becomes different and you get a consistent value for $T$.

Answer 2

Here is a nice, convergent series:

$$ \begin{align}1-\frac12+\frac13-\frac14+\dots & =\underbrace{1-\frac12-\frac14}_{1\text{ odd, }2\text{ evens}}+\frac13-\frac16-\frac18+\dots \\ & =\underbrace{\left(1-\frac12\right)-\frac14}_{1\text{ odd, }2\text{ evens?}}+\left(\frac13-\frac16\right)-\frac18+\dots \\ & =\frac12-\frac14+\frac16-\frac18+\dots\tag{calculate parenthesis?} \\ & =\frac12\left(1-\frac12+\frac13-\frac14+\dots \right) \end{align}$$

But wait a minute?!? Clearly the original series is larger than $0$,

$$\left(1-\frac12\right)+\left(\frac13-\frac14\right)+\dots>0?$$

But less than $1$,

$$1+\left(-\frac12+\frac13\right)+\left(-\frac14+\frac15\right)+\dots<1? $$

But we just showed above that

$$1-\frac12+\frac13-\frac14+\dots =\frac12\left(1-\frac12+\frac13-\frac14+\dots \right)?$$

So the sum exists and is equal to half of itself?

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Bonus points if you can figure out why.