Let $a_n>0$ for any positive integer $n$ and $\lim_{n \to \infty}a_{n} = a>0$. Prove that $\lim_{n \to \infty} \sqrt[n]{a_{n}} = 1$
My approach was to use Squeeze theorem, but I could not find the left-bound and right-bound. Thanks to Rudy the Reindeer for the solution. I overlooked the fact that the given limit is true for all $\varepsilon>0$, which allows me to take $\varepsilon$ that will give me the required bounds.
*edited, sorry for the missing $a_n$ in the given condition
*edit: Hi, I apologise if my question is not clear enough and I am still not sure what kind of questions are on-topic or off-topic since I just joined math.stackexchange recently. I tried my best to improve the question by explaining the difficulties that I face in solving this problem and I could not improve the question any further. Yet, I still do not know the specific reason on why my question is put on-hold. Thanks for the help! Have a great week ahead 🙂
Best Answer
Please state the problem correctly. The second expression is missing what you are taking the limit of, presumably $a_n$. Now for large $n, a_n$ is close to $a$, so $\sqrt[n]a_n$ is close to $\sqrt[n]a$