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I need to "study the limit behavior of the following sequence" and then compute the limit. The sequence is $a_n=\frac{n}{\sqrt[n]{n!}} $. This can also be written as $(\frac{n^n}{n!})^{(1/n)}$ I tried to prove it was monotonic because I felt that the sequence was increasing, and I ended up with a great mess, so I tried to prove the sequence was Cauchy and got stuck:

To prove the sequence is Cauchy, for $\epsilon>0,\exists N$ such that $m,n>N$ implies $|a_m-a_n|<\epsilon$ $$|a_m-a_n|=\bigg|\bigg(\frac{m^m}{m!}\bigg)^{1/m}-\bigg(\frac{n^n}{n!}\bigg)^{1/n}\bigg|\leq\bigg|\bigg(\frac{m^m}{m!}\bigg)^{1/m}\bigg|+\bigg|\bigg(\frac{n^n}{n!}\bigg)^{1/n}\bigg|=\bigg(\frac{m^m}{m!}\bigg)^{1/m}+\bigg(\frac{n^n}{n!}\bigg)^{1/n}=\frac{m}{m!^{1/m}}+\frac{n}{n!^{1/n}}\leq\frac{m}{m^{1/m}}+\frac{n}{n^{1/n}}<m+n$$ Which seems to be really wrong, except I don't know what else to do. Also I went on to compute the limit: Let $s_n=\frac{n^n}{n!}$ and recall that $\lim |\frac{s_{n+1}}{s_n}|=L=\lim|s_n|^{1/n}$ $$\bigg|\frac{s_{n+1}}{s_n}\bigg|=\bigg|\frac{(n+1)^{n+1}}{(n+1)!}*\frac{n!}{n^n}\bigg|=\frac{(n+1)^{n+1}}{(n+1)!}*\frac{n!}{n^n}=\bigg(\frac{n+1}{n}\bigg)^n=(1+\frac{1}{n})^n.$$ $$\lim|\frac{s_{n+1}}{s_n}|=\lim(1+\frac{1}{n})^n=e=L=\lim|s_n|^{1/n}=\lim a_n$$ Any help please?

user66807
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  • If you are deliberately using the fact that $|s_{n+1}/s_n| \to L$ implies $|s_n|^{1/n} \to L$, then you have already determined the limit behaviour of the sequence. Why do you still need to prove it's monotonic? – Erick Wong Mar 16 '13 at 01:06
  • I couldn't complete the monotonicity/Cauchy part of studying the behavior of the limit, so I just assumed it for now. So my question should say assuming that lim $|\frac{s_{n+1}}{s_n}|$ exists – user66807 Mar 16 '13 at 01:12
  • You have yet to prove that $\lim (1+1/n)^n$ exists, but you correctly stated it is equal to $e$? – Erick Wong Mar 16 '13 at 01:25
  • This was proved in class, so I think I'm allowed to use that without proving it. – user66807 Mar 16 '13 at 01:30
  • But that's the same as $\lim |\frac{s_{n+1}}{s_n}|$, isn't it? Perhaps you meant to say "assuming $\lim |s_n|^{1/n}$ exists". – Erick Wong Mar 16 '13 at 01:34
  • Well in the beginning I didn't know $\lim |\frac{s_{n+1}}{s_n}|$ was going to turn out to be $\lim (1+1/n)^n$ until I did the algebra, so I initially assumed that it's limit existed. After the algebra I realized that its limit had been proved in class, so I said that since $\lim |\frac{s_{n+1}}{s_n}|$ exists so does $\lim |s_n|^{1/n}$ and they are equal. – user66807 Mar 16 '13 at 01:43
  • I've seen this question many times. See also http://math.stackexchange.com/q/201906/9464 –  Mar 16 '13 at 02:50
  • @Jack yes I've seen it on here too and I examined the answers given as well. The aim of my question wasn't to compute the limit, that part I could do. I thought in order to answer the question I needed to prove the that the sequence was monotone/bounded or the Cauchy criterion, which is why I asked the question again. – user66807 Mar 16 '13 at 03:02

3 Answers3

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You might start with Stirling's approximation for the factorial.

Robert Israel
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An idea (Hint): define

$$a_n:=\frac{n!}{n^n}\Longrightarrow \frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1)^{n+1}}\frac{n^n}{n!}=\left(1+\frac{1}{n}\right)^{-n}\xrightarrow [n\to\infty]{} \frac{1}{e}<1$$

so by D'Alembert's test( = ratio test) , we get that the positive series

$$\sum_{n=1}^\infty \frac{n!}{n^n}\;\;\text{converges}\;\;\Longrightarrow\frac{n!}{n^n}\xrightarrow[n\to\infty]{}0$$

So that series is bounded....what can you then say about your series $\,\displaystyle{\frac{1}{\sqrt[n]{a_n}}}\,$ ...?

Another approach: Put

$$\frac{\sqrt[n]{n!}}{n}=\frac{1}{n}e^{\frac{1}{n}\log n!}=\frac{1}{n}e^{\frac{1}{n}\sum_{k=1}^n\log k}=e^{\frac{1}{n}\sum_{k=1}^n\log\frac{k}{n}}$$

But then we have a Riemann sum!:

$$\frac{1}{n}\sum_{k=1}^n\log\frac{k}{n}\xrightarrow[n\to\infty]{}\int\limits_0^1\log x\,dx=\left.\left(x\log x-x\right)\right|_0^1=-1\,\,(\text{note this is an improper integral...)}$$

So finally

$$\frac{\sqrt[n]{n!}}{n}\xrightarrow[n\to\infty]{}e^{-1}\Longrightarrow\frac{n}{\sqrt[n]{n!}}\xrightarrow[n\to\infty]{}e$$

DonAntonio
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  • Okay, I know that lim $\frac{a_{n+1}}{a_n}=\lim |a_n|^{1/n}$. Now 1 over this would give me $e>1$, but doesn't this mean my "series" is unbounded? Which would contradict the limit I found? Also, my question is about a sequence, since we haven't gone over series yet, would your hint still apply? – user66807 Mar 16 '13 at 01:08
  • Well, if you people haven't yet gone over series then I'm afraid the above is irrelevant, and probaly uncomprenhensible, for you. In a little while you shall reach this point, I guess. – DonAntonio Mar 16 '13 at 01:11
  • I understand some of it, and yes series are in the next chapter. Do you know any other way for me to go about it? Perhaps by proving it is Cauchy? – user66807 Mar 16 '13 at 01:15
  • I'm not sure...have you people studied Riemann Integral already? – DonAntonio Mar 16 '13 at 01:34
  • Nope, the basic things we learned are that a monotone/bounded sequence converges, if the sequence is Cauchy it converges. I think I was able to find the limit properly, but that was by making assumptions about the sequence. – user66807 Mar 16 '13 at 01:40
  • I see...and you say you haven't yet studied Stirling...right now I don't see any other approach. Sorry. – DonAntonio Mar 16 '13 at 01:45
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From the comments on the question, it seems like you have already answered this yourself without knowing it. You never used the fact that $|s_{n+1}/s_n|$ converges when verfiying that it is exactly equal to $(1+1/n)^n$. Since you know the latter converges, you know the exact same sequence converges without any prior assumption. Thus $|s_{n+1}/s_n| \to e$ and also $|a_n| \to e$ (since $a_n$ is positive, this also means $a_n \to e$).

Erick Wong
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  • I have a feeling that I am not understanding the "study the limit behavior of the sequence" properly. When I read this I feel that I have to first state that the sequence converges (which I only know how to do by proving monotonicity/boundedness or "Cauchy"). Any insight to what those directions are specifically asking me? And, should I say that: observe that $|s_{n+1}/s_n| = (1+1/n)^n$ which is known to converge to e, thus since $|s_{n+1}/s_n| \to e$ it follows that $|a_n| \to e$ ? – user66807 Mar 16 '13 at 02:02
  • @user66807 Yes, assuming all of those pieces have already been established, it's as simple as that (and nowhere near as complicated as the other answers). Any theorem that ends with "…then $a_n \to L$" is a possible tool for proving that $a_n$ converges, not just bounded monotonicity and the Cauchy criterion. Those two are useful for proving a sequence converges in contexts where the exact limit is hard to identify. – Erick Wong Mar 16 '13 at 02:27
  • That helps a lot! Thanks so much. I kept trying to show monotonocity and the Cauchy criterion and getting lost. – user66807 Mar 16 '13 at 02:31
  • @user66807 The weakest link is most likely the inference that if $|s_{n+1}/s_n| \to L$ then $|s_n|^{1/n} \to L$. This is a relatively subtle fact, and it's not clear whether you have learned it. For instance, it is not true in the reverse direction, so it is not fair to say that $\lim |s_{n+1}/s_n| = \lim |s_n|^{1/n}$ since that suggests a two-way equality. – Erick Wong Mar 16 '13 at 02:31
  • Oh thanks for pointing that out, I might have overlooked it. This is a Corollary in our textbook so I should be able to use it, and its proof is in the book as well. – user66807 Mar 16 '13 at 02:34