# inequality about $n^{1/n}$

When I was studying application of the sandwich theory, it was concluded that $n^{1/n}>1$ with regard to prove $\lim_{n\to\infty}n^{1/n}=1$.

In my opinion, it is obviously warranted that $n^{1/n}>0$. I think $n^{1/n}>1$ is certain only when $n>1$, but we can't be of the certain value of $n$. Does the $\lim_{n\to\infty}$ determine that $n^{1/n}>1$?

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 $n$ may be an integer. Whether it is or not, as it increases towards $+\infty$ it will exceed $1$ at which point $n^{1/n} \gt 1$. – Henry May 27 '12 at 10:22 Note that $(-1)^{1/(-1)} \lt 0$ – Henry May 27 '12 at 10:24 When $n\rightarrow\infty$,it is undoubted that $n>1$.But,meanwhile, $1/n\rightarrow 0$. Finally, we get cardinal number to be infinity and power exponent to be 0. Can we directly conclude $lim_{n\to\infty}n^{1/n} \rightarrow\ 1$? – Francis King May 27 '12 at 11:09 It is extremely dangerous to do a substitution of "$\infty$," whatever that may mean, for $n$. For example, $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{1/n}=e$. Mechanical substitution of "$\infty$" might suggest the answer is $1$. It isn't. – André Nicolas May 27 '12 at 14:31

If the notation $\lim_{n\to \infty}$ is used, it is understood that $n$ is a natural number. For $n=0$, the expression is undefined, and for $n=1$, it is trivially false, so yes, it is implicitly assumed that $n>1$.
Note that in the context of limits, $a_n>1$ is often only interesting for sufficiently large $n$.
 Why the notation $lim_{n/to/infty} implies that n is a natural number? Is this determined by the fact that the notation implys the limit to be sequence limit? – Francis King May 27 '12 at 10:49 If you don't specify the domain,$n$could be a real number or a natural number, but in case of ambiguity, one turns to convention, and the convention is that the letter$n$is most often used for natural numbers. – Phira May 27 '12 at 10:59 oh got it. I always confused limit of function and limit of sequence. In this case,can I conclude that$lim_{n\to\infty}$implies limit of sequence and that$lim_{x\to\infty}$implies limit of function? – Francis King May 27 '12 at 11:22 @FrancisKing Yes, in the absence of more context, you can conclude that this is meant. Obviously, it is not forbidden to call a real number$n$, so it might happen. – Phira May 27 '12 at 12:20 Thanks.Furthemore,I wonder why " for n=1, it is trivially false" and what exactly means for "trivially false"? – Francis King May 27 '12 at 13:24 show 1 more comment When we write mathematics, we usually stick to certain conventions. One is that$n$representas a natural number, i.e.$n\in \Bbb N$. Other variants can be$k$,$i$,$j$, but those are mostly used in sums, for the indices. So when one writes $$\lim_{n \to \infty} n^{\frac 1 n }$$ one is implicitly asking for the limit of the sequence$a_n = n^{\frac 1 n }$. It'd be very odd for someone to write$n$if they want$n$a real number. If$n$is real, we would usually use$y$,$x$,$r$,$q$, or specify that indeed it is real. Even so, the use of$n$would be avoided, since most of us have our brains already programmed after some time of math reading. Just for the sake of it, you will often see $$j,k,l,m,n \in \Bbb N$$ $$p,q,r \in \Bbb Q$$ $$a,b,c,d,x,y,z \in \Bbb R$$ $$x,y,z \in \Bbb C$$ I'm not saying this is strict (as for example Apostol uses$x$and$y\$ for integers, but he explicitly states that they are integers.) but it will help you when you write, for others to interpret what you're writing, and for you to read maths with more ease.