Mistakes in $\lim_{a\to \infty}(a^2 - a) = - \frac{1}{6}$?

Question

One can say, using Ramanujan summation or the zeta function regularization, that the sum $\sum_{k=1}^{\infty} k=- \frac{1}{12}$. Using this result I've gotten a very confusing and counterintuitive result, which I suspect it's wrong, but I can't see any mistakes in the reasoning. Could you tell me if it's right or not and where is the mistake? Consider a number $a \in N$, then it's square can be written as: $$a^2 = (a-1)^2 + 2(a-1) + 1$$ $$a^2 = (a-2)^2 + 2(a-2) + 1 + 2(a-1) + 1 $$ $$a^2 = (a-3)^2 + 2(a-3) + 1 + 2(a-2) + 1 + 2(a-1) + 1$$ $$...$$ $$a^2 = (a-a)^2 + 2(1+2+3+...+(a-1)) + a$$ $$a^2 = a + 2\sum_{k=1}^{a-1} k$$ Then: $$a^2 - a = 2\sum_{k=1}^{a-1} k$$ Taking limits on both sides: $$\lim_{a\to \infty}(a^2 - a) = \lim_{a\to \infty}2\sum_{k=1}^{a-1} k$$ Using the result from above we get: $$\lim_{a\to \infty}(a^2 - a) = 2(- \frac{1}{12}) = - \frac{1}{6}$$

Answer 1

While your working is admirable, your final step of 'taking limits' is completely undefined, i.e., ask yourself, what are you really doing when you 'take limits' in this context? What definition or process are you referring to?

If you are using the usual, elementary calculus notion of 'limit', then you have a divergent limit 'equalling' a divergent limit - it doesn't make sense to 'take' such a limit.

If you want to use (in)famous 'identity' $1+2+3+\ldots=-1/12$, then you have even more problems to worry about: firstly, it's vague as to what the symbol $\lim$ now means in your argument; secondly, it's not clear that, under whatever definition of $\lim$ you fix upon, $\lim_{a\rightarrow\infty}2\sum_{k=1}^{a-1} k=2\lim_{a\rightarrow\infty}\sum_{k=1}^{a-1} k$, and so you cannot necessarily conclude that $\lim_{a\rightarrow\infty}2\sum_{k=1}^{a-1} k=-1/6$.

That being said, I rather enjoyed your round-about way of proving the well-known formula $\sum_{k=1}^{a} k = a(a+1)/2$. It's just the 'taking limits' part that makes no sense.