Clearly this converges. I am asking whether it converges to $0$ or $1$ because both seem to make sense. Using LH, the sum becomes $$\lim_{n \to \infty}\frac{n}{n + \sum_{k=1}^{n}k}\Rightarrow _{LH}\lim_{n \to \infty}\frac{1}{1+0}=1$$ However I believe I am making a mistake because my idea is that since $\sum_{k=1}^{n}k$ is always a constant, I can differentiate term by term in terms of $d/dn$ to get $0$, but $\sum_{k=1}^{n}k$ is in terms of $k$. The case that this converges to $0$ is strong since $\sum_{k=1}^{n}k > n$ for all $n>1$, as in the denominator grows much faster than the numerator. Nonetheless, I want a solid understanding on why I cannot differentiate the sum term by term in terms of $d/dn$ and thus not use LH. Furthermore I suspect that we can just convert $\sum_{k=1}^{n}k$ to its sequence of partial sums which would be in terms of $n$ and then evaluate the limit from there, but I got lost trying to do this.
2026-05-04 18:13:53.1777918433
Does $\lim_{n \to \infty}\frac{n}{n + \sum_{k=1}^{n}k}$ converge to $0$ or $1$?
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You can write $$ \frac{n}{n+\sum_{k=1}^nk}=\frac1{1+\frac1n\,\sum_{k=1}^nk}. $$ Now, $$ \frac1n\,\sum_{k=1}^nk=\frac1n\,\frac{n(n+1)}2=\frac{n+1}2. $$ Thus $$ \frac{n}{n+\sum_{k=1}^nk}=\frac1{1+\frac{n+1}2}\xrightarrow[n\to\infty]{} 0. $$
It doesn't make sense to apply L'Hôpital, as you don't have functions over the real line, just discrete functions. In particular you cannot define the sum for all $x$.