Find the limit of $\lim_{n \to \infty} \frac{\sqrt[n]{(n+2025)!}}{n - 2025}$

Question

my calculus-1 professor gave us this as homework to find the limit above when n goes to infinity. I tried to put the n square in the bottom but that didn't solve anything. I tried a few approaches:

Stirling approximation: Using I approximated the numerator and then looked at the overall expression, but it became quite messy and hard to evaluate precisely.

Taking logarithms: I considered: but this also got complicated and didn't directly help compute the original limit.

I have been told this limit is best approached using Stolz's theorem in an indirect way, but I'm not entirely sure how to restructure the limit to apply it correctly.

Can anyone help show how to apply Stolz's indirectly here? In particular, what substitution or reformulation would make it possible to use the theorem effectively?

Answer 1

We can use that

• Do the sequences from the ratio and root tests converge to the same limit?

with

$$\frac{a_{n+1}}{a_n}=\frac{(n+1+2025)!}{(n+1 - 2025)^{n+1}}\frac{(n - 2025)^n}{(n+2025)!}=\frac{n+2026}{n-2024}\left(1-\frac1{n-2024}\right)^n$$

Answer 2

I suppose that $2025$ is a red herring.

Define $$f(a)= \frac{\sqrt[n]{(n+a)!}}{n - a} $$ Take logarithms $$\log(f(a)=\frac 1n \log((n+a)!)-\log(n-a)$$ Use Stirling approximation $$\log(f(a))=-1+\frac{(2 a+1) \log (n)}{2 n}+\frac{2 a+\log (2 \pi )}{2 n}+O\left(\frac{1}{n^2}\right)$$ What is left when $n\to \infty$ ?

Answer 3

Note $$\begin{eqnarray} &&\lim_{n \to \infty}\ln\bigg(\frac{\sqrt[n]{(n+a)!}}{n-a}\bigg)=\lim_{n \to \infty}\ln\bigg(\frac{\sqrt[n]{(n+a)!}}{n}\bigg)\\ &=&\lim_{n \to \infty}\sum_{k=1}^{n+a}\frac{\ln(k)}{n}-\sum_{k=1}^{a}\frac{\ln(k)}{n}-\ln(n)=\lim_{n \to \infty}\frac{\sum_{k=1}^{n+a}\ln(k)-n\ln(n)}{n}\\ &\overset{Stolz' Theorem}=&\lim_{n \to \infty}\frac{\bigg[\sum_{k=1}^{n+1+a}\ln(k)-(n+1)\ln(n+1)\bigg]-\bigg[\sum_{k=1}^{n+a}\ln(k)-n\ln(n)\bigg]}{(n+1)-n}\\ &=&\lim_{n \to \infty}\bigg[\ln(n+1+a)-(n+1)\ln(n+1)+n\ln(n)\bigg]\\ &=&\lim_{n \to \infty}\bigg[\ln n+\ln(1+\frac{a}n)-(n+1)\ln n-(n+1)\ln(1+\frac1n)+n\ln n\bigg]\\ &=&\lim_{n \to \infty}\bigg[\ln(1+\frac{2025}n)-(n+1)\ln(1+\frac1n)\bigg]=-1 \end{eqnarray}$$ and hence $$\lim_{n \to \infty} \frac{\sqrt[n]{(n+2025)!}}{n - 2025}=\frac1e.$$

Answer 4

Similar to xpaul's idea.

The limit in OP is the same as $\lim_{n\infty}\frac{\sqrt[n]{n!}}{n}$. Taking $\ln$ of the limit function, we need to find $$\lim_{n\to\infty}\frac1n\ln(n!)-n\\ =\lim_{n\to\infty}\frac1n\sum_{k=1}^n\ln(\tfrac kn)$$ which is $$\int_0^1\ln xdx=-1.$$