Summation of natural numbers divided by summation of even numbers

Question

I came across this equation (which I think is not correct):

$$ \frac{1 + 2 + 3 + 4 + ...}{2 + 4 + 6 + 8 + ...} = \frac{1}{2} $$

One would argue that above equation can be simplified to $\frac{\sum_{i=1}^{n=\infty} i}{2\sum_{i=1}^{n=\infty} i}$ and therefore it's valid.

But $\lim\limits_{n\to\infty} \sum\limits_{i=1}^n i$ is a divergent series. I also know that there is same number of even numbers as natural numbers, $\Bbb N_0$. But that equation basically says that there are twice as many even numbers as natural numbers.

How would you invalidate the aforementioned equation considering points above?

Answer 1

There is a subtlety here.

$\frac{1+2+3+...+n}{2+4+6+...+2n} = \frac {\frac 12 (n)(n+1)}{n(n+1)} = \frac 12$

No argument, I hope?

And actually, using the same argument:

$\lim_{n\to \infty}\frac{1+2+3+...+n}{2+4+6+...+2n} = \lim_{n\to \infty}\frac {\frac 12 (n)(n+1)}{n(n+1)}=\frac 12$

Because that is computing the limit of the ratio taken as a whole, and this limit exists and it is $\frac 12$.

But if you write the expression as either $\frac{1+2+3+...}{2+4+6+...}$ or $\frac{\lim_{n\to \infty}(1+2+3+...+n)}{\lim_{n\to \infty}(2+4+6+...+2n)}$, which are equivalent formulations with the latter being more precise - then the limit does not exist because both the top and bottom are divergent series.

Answer 2

For some infinite series ($a_1, a_2, a_3,\dots)$, we can define the sum $\sum\limits_{i=1}^\infty a_i$ as the limit $\lim\limits_{n\to\infty} \sum\limits_{i=1}^n a_i$.

In the case where $a_i=i$, however, $\sum\limits_{i=1}^n i=\frac{n(n+1)}2,$ and the limit does not converge,

so the infinite sum $\sum\limits_{i=1}^\infty i$ is not defined, so the equation does not make sense.