Determine $\lim_{n\to\infty} \left(\left(3\sqrt{n}\right)^{\frac{1}{2n}}\right)$

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!

• Do you mean $\lim_{n\rightarrow\infty}$? – Larry Dec 4 '18 at 4:12
• Yes, I will edit. – user573025 Dec 4 '18 at 4:13

2 Answers

Fist of all you shoul write $$\left(3\sqrt{n}\right)^{\frac{1}{2n}}=1+k_n$$.

$$n=\sqrt{\left(\frac{1}{3}\left(1+k\right)^{2n}\right)}$$ is not right.

We get

$$(1+k_n)^{2n}=3 \sqrt{n}$$. Use Bernoulli to get $$3 \sqrt{n} \ge 1+2n k_n \ge 2n k_n$$.

This gives

$$0 \le k_n \le \frac{3}{2 \sqrt{n}}$$ which shows that $$k_n \to 0$$.

A easier proof:

$$(3\sqrt{n})^{\frac{1}{2n}}=( \sqrt{3})^{1/n}\cdot (n^{1/n})^{1/4} \to 1.$$

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 $$k\to\infty$$.