# Why is $\frac{\sum_{n=1}^{\infty} n}{\sum_{n=1}^{\infty} n}$ indeterminate?

We all know that $$\dfrac{f(x)}{f(x)} = 1$$ (where $$f(x) \neq 0$$) and that $$\sum_{n=1}^{x} n = \dfrac{x(x+1)}{2}$$.

So, given $$f(x) \stackrel{\text{def}}{=} \sum_{n=1}^{x} n$$, we show that $$\dfrac{f(x)}{f(x)} = \dfrac{\frac{x(x+1)}{2}}{\frac{x(x+1)}{2}} = \dfrac{x(x+1)}{x(x+1)} = 1$$ (where $$x \not\in \{-1, 0\}$$).

From this, it seems logical that $$\dfrac{f(\infty)}{f(\infty)}$$ would equal $$1$$. Now, before you bash me for using $$\infty$$ like a number, I know $$\infty$$ isn't a number and can't be used as one, but bear with me. However, WolframAlpha begs to differ and spits out $$(indeterminate)$$. I assume it's calculating $$\dfrac{\frac{\infty(\infty+1)}{2}}{\frac{\infty(\infty+1)}{2}} = \dfrac{\infty(\infty+1)}{\infty(\infty+1)} = \dfrac{\infty}{\infty}$$ which is $$(indeterminate)$$.

All that makes sense, but because $$\infty$$ isn't a number, you can't calculate $$f(\infty)$$ and (from what I've been taught) instead must calculate $$\lim_{x \to \infty} \dfrac{f(x)}{f(x)}$$, which works out as long as $$x \not\in \{-1, 0\}$$:

$$\dfrac{f(1)}{f(1)} = \dfrac{1}{1} = 1$$

$$\dfrac{f(2)}{f(2)} = \dfrac{3}{3} = 1$$

$${}\qquad\vdots$$

$$\dfrac{f(10^{10})}{f(10^{10})} = \dfrac{50\space000\space000\space005\space000\space000\space000}{50\space000\space000\space005\space000\space000\space000} = 1$$

And, of course, it works out to be $$1$$ as long as $$x \not\in \{-1, 0\}$$. In addition, when graphed as $$\dfrac{\frac{x(x+1)}{2}}{\frac{x(x+1)}{2}}$$ (WolframAlpha doesn't like the sum form), you get none other than a $$y = 1$$ plot (with holes at $$x \in \{-1, 0\}$$):

What gives? Is WolframAlpha wrong again, or have I just been taught incorrectly again (like how $$\sqrt{x^2} = x$$)?

If I wanted to use the analytically continued Riemann-Zeta function, I could use $$\zeta(-1)$$ instead of $$f(\infty)$$, I get $$\dfrac{\zeta(-1)}{\zeta(-1)} = \dfrac{-\frac{1}{12}}{-\frac{1}{12}} = 1$$. But this is out of the scope of the question.

Sure,

$$\lim_{x\to\infty}\frac{\displaystyle\sum_{n=1}^xn}{\displaystyle\sum_{n=1}^xn}=1.$$

However, $\displaystyle\sum_{n=1}^\infty n$ is defined to be $\displaystyle\lim_{x\to\infty}\sum_{n=1}^x n$, which does not exist, hence

$$\frac{\displaystyle \sum_{n=1}^\infty n}{\displaystyle \sum_{n=1}^\infty n} =\frac{\displaystyle \lim_{x\to\infty}\sum_{n=1}^x n}{\displaystyle \lim_{x\to\infty}\sum_{n=1}^x n}$$

is a ratio of two things that do not exist. So, of course, the ratio does not exist.

There are strict rules that tell us when it is okay to pull $\lim$s out of expressions, or to consolidate multiple $\lim$s together, and you have been ignoring these rules.

The numerator and denominator are $\infty$, so you're asking wolfram alpha "What is $\infty/\infty?$" which doesn't have an answer. If instead you asked the different question: What is $\lim_{N\rightarrow\infty} \sum_0^Nn / \sum_0^Nn$? you would get 1.

$+\infty$ is a number, can be used as a number (although the extended real numbers don't satisfy all of the algebraic identities that the real numbers do), and $(+\infty)/(+\infty)$ is undefined (in the same way that $0/0$ is undefined).

And $\sum_{n=1}^{\inf} n = +\infty$.

The expression $\frac{\sum_{n=1}^{\inf} n}{\sum_{n=1}^{\inf} n}$ is not indeterminate: it is undefined.