Given a sequence of real numbers $r$ such that $\lim_{n\to\infty}r_n^\frac{1}{n}=1$, then what can we say about $\lim_{n\to\infty}r_n$?

If $\lim_{n\to\infty}r_n^\frac{1}{n}=0.99$, then $r_n=0.99^n$, so $\lim_{n\to\infty}r_n=0$.

And if $\lim_{n\to\infty}r_n^\frac{1}{n}=1.01$, then $r_n=1.01^n$, so $\lim_{n\to\infty}r_n=\infty$

But in the case of $\lim_{n\to\infty}r_n^\frac{1}{n} = 1$, I am not sure what can be concluded about $\lim_{n\to\infty}r_n$

  • $\begingroup$ You say that "$r$ is a real number" then how it then becomes $r=0.99^n$ which is not a number but a sequence? $\endgroup$
    – A.Γ.
    Commented Oct 10, 2015 at 19:19
  • $\begingroup$ @A.G. Thanks for correcting. $r$ is a sequence of real numbers. $\endgroup$ Commented Oct 10, 2015 at 19:21
  • $\begingroup$ @Aniket The notation $r^{\frac{1}{n}}$ is very confusing, since you seem in other expressions to be using $n$ to index $r$, but there you are using it as an exponent. What exactly do you mean by it? Do you mean $r_n^\frac{1}{n}$? $\endgroup$ Commented Oct 10, 2015 at 19:25

2 Answers 2


Nothing can be said about $\lim_{n\to\infty}r_n$: it may be $+\infty$ as for $r_n=n$, it may be zero as for $r_n=\frac{1}{n}$, it may be any number between as for $r_n=r\in(0,+\infty)$ or it may not exist as for $r_n=2+\sin(n)$.

P.S. To handle the limits above rewrite $r_n^{\frac{1}{n}}=e^{\frac{1}{n}\ln r_n}$ and see that $\lim_{n\to+\infty}\frac{1}{n}\ln r_n=0$.

EDIT: To make it clearer let us consider the following class of sequences $r_n$ $$ \frac{c}{n}\le r_n\le C\cdot n $$ with any positive constants $c$ and $C$. It is quite a large class, since $r_n$ can vary from going to zero (as $1/n$) to going to infinity (as $n$), with any limit in between being possible. There are also oscillating sequences in the class that have no limit. Now, take the logarithm of this inequality and divide by $n$ $$ \ln c -\ln n\le \ln r_n\le \ln C+\ln n\qquad\Rightarrow\qquad \underbrace{\frac{\ln c}{n} -\frac{\ln n}{n}}_{\to 0}\le \frac{\ln r_n}{n}\le \underbrace{\frac{\ln C}{n}+\frac{\ln n}{n}}_{\to 0}. $$ Since both sides go to zero as $n\to\infty$, by the squeeze theorem we get that the middle part goes to zero as well, therefore $$ \lim_{n\to\infty}r_n^{\frac1n}=\lim_{n\to\infty}e^{\frac1n\ln r_n}=e^{\lim_{n\to\infty}\frac1n\ln r_n}=1. $$ Conclusion: the limit of $n^\text{th}$ root is equal to one for a huge class of sequences $r_n$ that may themselves have any limit between $0$ and $+\infty$ or none.

  • $\begingroup$ Concise and complete. Excellent!! +1 $\endgroup$
    – Mark Viola
    Commented Oct 10, 2015 at 20:43
  • $\begingroup$ Please show which oscillating sequences you are talking about. $\endgroup$ Commented Oct 22, 2015 at 15:54
  • $\begingroup$ @Aniket The one I have mentioned above is $r_n=2+\sin n$. Another even simpler is $1,2,1,2,1,2,\ldots$ $\endgroup$
    – A.Γ.
    Commented Oct 22, 2015 at 20:09

You actually cannot say about sequences with limit 1, since if we consider x(n)=n, and y(n) = 1/n, we have n^(1/n) and (1/(n^(1/n))) converging to 1. Proof: Consider z(n) = n^(1/n). Define b(n) + 1 = z(n), so n= (1 + b(n))^n, so by the expansion of (a+b)^n, we have (1 + b(n))^n > [n(n-1)/2]*(b(n)^2), so 2/(n-1) > (b(n)^2)>0, so by Archimedian Property, {2/n-1} converges to 0, {0} converges to 0, so by the Sandwich Lemma, we have {b(n)^2} converges to 0, so {b(n)} converges to 0 (Easy to show). Therefore, z(n) converges to 1+0 = 1, by Algebra of Convergent Sequences. Also, w(n)= y(n)^(1/n) now converges to 1/1=1. Therefore, while both obey the hypothesis, x(n) diverges (N is unbounded above), while y(n) converges to 0, by the Archimedian Property.

  • $\begingroup$ y(n) is a divergent sequence. Check the use of words. It does not converge to 0, the terms tend to 0 as $n\to\infty$ $\endgroup$ Commented Oct 31, 2015 at 11:21
  • $\begingroup$ @Aniket: Please note that y(n) = 1/n. Thus, y(n) will, as you say, "tend to 0 as n tends to infinity". Therefore, y(n) is NOT divergent. Perhaps, you are confusing it with the limit of the partial sums of {1/n}, that is the harmonic series, which diverges. Please explain to me why exactly I should not say y(n) converges to 0. $\endgroup$
    – user270212
    Commented Nov 12, 2015 at 20:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .