Determine $\lim_{n\to\infty} \left(\left(3\sqrt{n}\right)^{\frac{1}{2n}}\right)$
Is there a way to determine this limit without using the properties of the logarithm function? Anyways, I am not sure how to determine this limit so I claim the limit is $1$ by its similarity to the $\lim n^{\frac{1}{n}}$ so I set out to prove this:
Proof:
We need to find some $K\in\mathbb{N}$ such that $\forall n\ge K,$ we have that $\left|\left(3\sqrt{n}\right)^{\frac{1}{2n}}-1\right|<\epsilon$ for all $\epsilon>0$. Observe that $\left(3\sqrt{n}\right)^{\frac{1}{2n}}>1 \space\forall n$ so we can write $\left(3\sqrt{n}\right)^{\frac{1}{2n}}=1+k$ for some $k>0.$ Then we have that $$n=\sqrt{\left(\frac{1}{3}\left(1+k\right)^{2n}\right)}$$ I am stuck here. I think the next step is to apply the binomial theorem but I’m not sure how to deal with the $4n$ exactly. And maybe this method doesn't work at all. Any pointers would be great. Thanks!
Best Answer
Your limit really is $$\lim \left(\left(3\sqrt{n}\right)^{\frac{1}{2n}}\right)=\lim \left(\left(9n\right)^{\frac{1}{9n}}\right)^\frac94=\left(\lim_{k\to\infty} k^\frac1k\right)^{9/4}$$ and for $\lim_{k\to\infty} k^\frac1k=1$ one may use conventional proof, that is when $k>1$ let $k^\frac1k=1+A_k$ then with Binomial theorem $$k=(1+A_k)^k=1+kA_k+\dfrac12k(k-1)A_k^2+\cdots>\dfrac12k(k-1)A_k^2$$ which shows $$A_k<\sqrt{\dfrac{2}{k-1}}\to0$$ as $k\to\infty$.