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?
Best Answer
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.