Correct usage of Stolz-Cesaro theorem in finding a limit

Question

Is this solution correct?

I've seen this solution before, but I don't know why it is so. And which rule?

The task is to find: $\omega=\displaystyle\lim_{n\to +\infty}\left(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right)$

I know that this $\lim$ equals $\frac{1}{e},$ but do you see if this solution is correct or not?

Solution is :

$\displaystyle\lim_{n\to +\infty}\left(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right)=\displaystyle\lim_{n\to +\infty}\left((n+1)\sqrt[n+1]{\frac{(n+1)!}{(n+1)^{n+1}}}-n\sqrt[n]{\frac{n!}{n^{n}}}\right)$

$=\displaystyle\lim_{n\to +\infty}\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^{n}}}$

$=\frac{1}{e}$

Notes & suggestions of mine:

I think he uses Stolz-Cesaro theorem:

See that

$\displaystyle\lim_{n\to +\infty}\frac{a_{n}}{b_{n}}=\displaystyle\lim_{n\to +\infty}\frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}$

Chosen $a_{n}=n\sqrt[n]{\frac{n!}{n^n}}$

and $b_{n}=n$ clearly $b_{n}$ goes to $+\infty$ and $a_{n}$ goes to $+\infty$ because $n.\frac{1}{e}=+\infty$

we obtain:

$\omega=\displaystyle\lim_{n\to +\infty}\frac{n\sqrt[n]{n!}}{n}$

Then use Cauchy-d'Alembert

$=\displaystyle\lim_{n\to +\infty}\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^{n}}}=\frac{1}{e}$

Is my explanation O.K.? May I ask for a correction?

Any remark ?

Yes, that is how that argument can go using Stolz's. – MoonLightSyzygy Dec 22 '19 at 18:11 — MoonLightSyzygy, Dec 22 '19 at 18:11
See https://math.stackexchange.com/q/2020153/72031 – Paramanand Singh Dec 23 '19 at 09:47 — Paramanand Singh, Dec 23 '19 at 09:47

score 2 · Answer 1 · answered Dec 23 '19 at 06:58

2

The use of Cesaro-Stolz as you mention is invalid. In application of Cesaro-Stolz the existence of limit of $\dfrac{a_{n+1}-a_n}{b_{n+1}-b_n}$ is a hypotheses and not the conclusion. Thus you have tried to use Cesaro-Stolz in reverse which does not work.

answered Dec 23 '19 at 06:58

Paramanand Singh

87,309

Why , what about reserve stolz-Cesaro ? – Ellen Ellen Dec 23 '19 at 11:35
@EllenEllen: converse of that theorem does not hold in general and therefore your approach is wrong. – Paramanand Singh Dec 23 '19 at 11:38
But why i use it this method in many limits and I get always correct answer !! – Ellen Ellen Dec 23 '19 at 13:56
@EllenEllen: Here is a fact from rules of logic: a wrong approach does not guarantee wrong answer. Hence don't be charmed by the illusion of wrong approaches. They may fail when you need them most. – Paramanand Singh Dec 23 '19 at 14:11
@EllenEllen : Cesaro-Stolz is one of my favorite tools but it needs to be used in proper manner. – Paramanand Singh Dec 23 '19 at 14:11
@EllenEllen : your use of Cauchy-d'Alembert at the end is correct. Also note that Cauchy-d'Alembert follows from Cesaro-Stolz. – Paramanand Singh Dec 23 '19 at 14:15
ok thanks so this solution wrong ! , but see that $B\implies A$ sometimes we see if $A$ correct then B also correct ? So I think here we see if $\lim\frac{a_n}{b_n}=L$ then $\lim\frac{a_{n+1}-a_n}{b_{n+1}-b_{n}}=L$ ? – Ellen Ellen Dec 23 '19 at 14:18

PinkyWay · Answer 2 · 2019-12-23T08:56:10.303

-1

I don't think Stolz-Bolzano theorem is necessary here (: , but I agree with everything you explained. Nevertheless, I'll write down everything in detail for everyone who might have not understood similar proofs to other (similar) posts.

Since $n\to\infty$, $n+1$ behaves (more or less) similarly:

Inductively: $$\lim_{n\to\infty}{\sqrt[n+1]{\frac{(n+1)!}{(n+1)^{n+1}}}}=\lim_{n\to\infty}{\sqrt[n]{\frac{n!}{n^n}}}$$ $$\implies\lim_{n\to +\infty}\left(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right)=\lim_{n\to +\infty}\left((n+1)\sqrt[n+1]{\frac{(n+1)!}{(n+1)^{n+1}}}-n\sqrt[n]{\frac{n!}{n^{n}}}\right)=\lim_{n\to\infty}\Bigg((n+1)\sqrt[n]{\frac{n!}{n^n}}-n\sqrt[n]{\frac{n!}{n^n}}\Bigg)=\lim_{n\to\infty}{\sqrt[n]{\frac{n!}{n^n}}}=\lim_{n\to\infty}\frac{n!}{n^n}$$ As proven here: Why is $\lim\limits_{n \to +\infty }{\sqrt[n]{a_1 a_2 \ldots a_n}} =\lim\limits_{n \to +\infty}{a_n}$

You can apply the Cauchy-D'Alembert criterion already here:$$\lim_{n\to\infty}\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}}=\lim_{n\to\infty}\frac{n^n(n+1)!}{(n+1)^{n+1}n!}=\lim_{n\to\infty}\frac{n^n(n+1)}{(n+1)^{n+1}}=\lim_{n\to\infty}\frac{n^n}{(n+1)^n}=\lim_{x\to\infty}\frac{1}{\Big(\frac{n+1}{n}\Big)^n}=\lim_{n\to\infty}\frac{1}{\Big(1+\frac{1}{n}\Big)^n}=\frac{1}{e}$$

edited Dec 23 '19 at 08:56

answered Dec 22 '19 at 22:15

PinkyWay

4,565

Your approach is wrong. You can't replace one expression by another in this manner. - 1 – Paramanand Singh Dec 23 '19 at 07:01
@ParamanandSingh thank you for the remark. May I ask if this kind of solution is allowed at all in some cases? I thought I could do this since I had just a term of a sequence, not sum or product. I appreciate your comment because I also learn from that and benefit from it and gain more experience as a first-year student. – PinkyWay Dec 23 '19 at 10:24
1

Ok one can't leave the fundamentals. Thus you can't replace $A$ by $B$ unless $A=B$. I have provided some details in this answer which allows for certain manipulation while evaluating limits: https://math.stackexchange.com/a/1783818/72031 – Paramanand Singh Dec 23 '19 at 10:29
@ParamanandSingh, thank you very much for help and understanding! – PinkyWay Dec 23 '19 at 10:31

Correct usage of Stolz-Cesaro theorem in finding a limit

2 Answers2