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How to prove $$\lim_{n \rightarrow \infty}\sqrt[n]{n}=1.$$

I have problem in proving this statement at the beginning my textbook says:

Suppose $f_{n}=\sqrt[n]{n}=1+h_{n}$ where does this $1+h_{n}$ come from? and what does it mean? if we can write $\sqrt[n]{n}=1+h_{n}$ then $f_{n}=1+h_{n}$ so by defintion of convergence let $\varepsilon\gt0$ be given. There exists a positive integer $K\gt\frac{1}{h_{n}}$ such that $|f_{n}-l|\lt\varepsilon \implies |1+h_{n}-1|\lt\varepsilon\implies h_{n}\lt\varepsilon$...therefore $\sqrt[n]{n}\to 1$....Am i right?


marked as duplicate by Community Apr 10 '16 at 20:35

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  • $\begingroup$ Any number can be written in the form $1+h$. I guess the textbook wants you to write $\sqrt[n]{n}$ that way to see what you can learn about $h$ (named $h_n$ here, since it depends on $n$). Have you tried taking the $n$th power on both sides of of the stated equation? $\endgroup$ – Harald Hanche-Olsen Apr 10 '16 at 20:25
  • $\begingroup$ They intend for you to show that $h_n$ tends to zero... $\endgroup$ – KR136 Apr 10 '16 at 20:29
  • $\begingroup$ @SubhadeepDey I never meant to question that, though perhaps my comment made it sound that way. Let me edit it. $\endgroup$ – KR136 Apr 10 '16 at 20:33
  • $\begingroup$ sorry to say but I m still confuse with 1+h_{n}..can any one please explain what is meant by 1+h_{n} when we expand \sqrt[n]{n} $\endgroup$ – T.Noor Apr 10 '16 at 20:45
  • $\begingroup$ My first comment did not explain it adequately? Why is that so? $\endgroup$ – Harald Hanche-Olsen Apr 10 '16 at 20:51

I prefer another approach. Notice that $$ \log\lim_{n\to\infty}\sqrt[n]n=\lim_{n\to\infty}\frac{\log n}{n}=\lim_{x\to\infty}\frac{\log x}{x}=\lim_{x\to\infty}\frac{1}{x}=0, $$ passing to $\mathbb R$ and using L'Hôpital's rule. Therefore, $$ \lim_{n\to\infty}\sqrt[n]n=\exp\log\lim_{n\to\infty}\sqrt[n]n=e^0=1. $$

  • $\begingroup$ Yes, but since this is at the beginning of a textbook, it is possible that L'Hôpital isn't know yet. Or the natural logarithm. $\endgroup$ – Harald Hanche-Olsen Apr 10 '16 at 20:26
  • $\begingroup$ Most obviously only the OP can say, so either you provide an answer or let the OP clarify anything that needs to be clarified. PS: Honestly, I would like to see other proofs. $\endgroup$ – John B Apr 10 '16 at 20:30
  • $\begingroup$ There are other proofs in the posts linked in the comments to the question. $\endgroup$ – Harald Hanche-Olsen Apr 10 '16 at 20:33
  • $\begingroup$ yes i know this way to prove it.. $\endgroup$ – T.Noor Apr 10 '16 at 20:34
  • $\begingroup$ My second comment was a response to your remark that you would like to see other proofs. And yes, I am aware of the timing. Sorry if I have offended you, that was not my intention. But the OP clearly stated that the problem was stated at the beginning of a textbook, which indicates that not a lot of theory may be assumed. And he specifically asked about $1+h_n$, which you did not address. My initial comment was only intended to point this out. Your answer is of course correct and quite good, but seems a bit off the mark for the above reasons. But not enough off the mark for a flame war. 8-) $\endgroup$ – Harald Hanche-Olsen Apr 10 '16 at 20:48


$$\sqrt[n]{n}=e^{\frac 1n \log(n)}$$


$$\log(n)\le \frac{n^{\alpha}-1}{\alpha}$$

for all $\alpha>0$.


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