# L'hopital's rule discrepancy [closed]

The value of

$$\lim_{n\to\infty} \frac{1 + 2 + 3 + \dots + n}{n^2 + 100}$$

In this question, why can't we use L'Hopital's rule straight away?

• Is that supposed to be $n^2$ in the denominator? May 15 at 15:40
• Also, how do you intend to differentiate $1+2+\dots+n$ with respect to $n$? This is not a function on the real line May 15 at 15:41
• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
– Community Bot
May 15 at 15:42
• You can use a kind of l'Hospital here. Write $f(n)$ for the numerator and $g(n)$ for the denominator. We $f(n+1)-f(n)=n+1$ and $g(n+1)-g(n)=2n+1$. Such differences are a working replacement for the derivative (assuming you don't have a useful formula for $f$). This would lead to a limit of $1/2$, which is correct :-) May 15 at 16:18
• Recall that $1 + 2 + 3 + \dots + n =\frac{n(n+1)}2$
– user
May 15 at 17:11

1. There is a discrete version of L'Hôpital's rule: the Stolz–Cesàro theorem, that is (under similar conditions as L'Hôpital's rule):

$$\lim_{n\rightarrow \infty}\frac{f_n}{g_n}=\lim_{n\rightarrow\infty}\frac{f_{n+1}-f_n}{g_{n+1}-g_n}$$

1. It's possible to use L'Hôpital's, but you have to extend the function $$f(n), g(n)$$ to be functions $$f(x), g(x)$$ defined over $$\mathbb R$$ (or at least for sufficiently large $$x$$). This is straightforward for $$g(n)=n^2+100$$ that can be extended as $$g(x)=x^2+100$$, but not so much for $$f(n)=1+2+\cdots+n$$, unless you recognize $$f(n)=\frac{n(n+1)}{2}$$, then $$f(x)=\frac{x(x+1)}{2}$$. In theory, the extension is not unique and it's possible that the limit exists for discrete $$n$$, but not for continuous $$x$$ if a bad extension is chosen, therefore the extension matters. But in practice, these extensions are usually very natural, though it's not always easy, take e.g. Stirling's approximation: $$\lim_{n\rightarrow} \frac{n!}{\sqrt{2\pi n}(\frac{n}{e})^n} = 1$$ where $$n!$$ can be extended to the Gamma function but it's far less obvoius how to do so. (I don't know if either the L'Hôpital's or the Stolz–Cesàro can be used to establish Stirling, probably not, just to illustrate the point of extending the function.)
• Can I not differentiate 1+2+3+4...+ n to be 1, taking constants as zero ? May 16 at 6:38
• In $1+2+3+\cdots+(n-3)+(n-2)+(n-1)+n$, if $1, 2, 3$ are constants, but $n-3, n-2, n-1, n$ are not. When the term stopped being a constant? May 16 at 6:54
• @Justauser about halfway /hj May 16 at 9:40
• @ronno good one, and you just explained where the $1/2$ in the derivative comes from, but there is still a whole $n$ to be found! (/hj as well). May 16 at 10:13
• No, you have two terms: $(0 + 0 + \dots + 1)$ which gives an $n/2$ from half of the terms increasing with $n$, and the average of $(1 + 2 + \dots + n)$ from the number of terms increasing, which gives you an $(n + 1)/2$. Adding up you get $n + 1/2$, which is the correct derivative. May 16 at 12:43