3
$\begingroup$

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 $x \in \{-1, 0\}$):

plot

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.

$\endgroup$
1
9
$\begingroup$

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.

$\endgroup$
0
1
$\begingroup$

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.

$\endgroup$
0
1
$\begingroup$

$+\infty$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.