# Is $\lim \limits_{n\to\infty}\frac{\sum\limits_{r=1}^{n}r^0}{\sum \limits_{r=1}^{n}r} =6$?

$$\frac{\sum\limits_{r=1}^{\infty}r^0}{\sum\limits_{r=1}^{\infty}r}$$

Apparently, in special circumstances, the denominator of the above fraction can be considered equal to $-1/12$ and the numerator can be considered to equal $-1/2$.

If I use those values, I get $\large\frac{(\frac{-1}{2})}{(\frac{-1}{12})}=6$

I'm wondering if this would also be the correct answer for $$\lim_{n\to\infty}\frac{\sum\limits_{r=1}^{n}r^0}{\sum\limits_{r=1}^{n}r}$$

When people write $\sum_{r=1}^{\infty}r=\frac{-1}{12}$ or $\sum_{r=1}^{\infty} r^0=\frac{-1}{2}$, it is purely from a severe abuse of notation. These are divergent series; they do not converge. However, the Riemann zeta function, which is the analytic continuation of the series $\sum_{n=1}^{\infty}\frac{1}{n^s}$ (where $s$ has real part greater than $1$) is sometimes used to assign values to these things. This is entirely a notational thing, however, and we are essentially declaring the nonsense statement $\sum_{r=1}^{\infty}r=\zeta(-1)$.
Also, note that $\frac{\sum_{r=1}^{n}r^0}{\sum_{r=1}^{n}r}=\frac{n}{n(n+1)/2}=\frac{2}{n+1}$ converges to $0$ as $n\to\infty$.
The answer is $0$, $$=\lim_{n \to \infty} \frac{\sum_{r=1}^n 1}{\sum_{r=1}^n r}$$ $$=\lim_{n \to \infty} \frac{2n}{n(n+1)}$$ $$=0$$