Better way to do $\lim_{n\to\infty}\frac{(3n-1)^{1/n}}{{\sqrt{2}}^{1/n}}$

Question

I want to find this limit:

$$\lim_{n\to\infty}\frac{(3n-1)^{1/n}}{{\sqrt{2}}}$$

because im trying to prove the convergence of the series

$$\sum_{n=1}^{\infty}\frac{3n-1}{(\sqrt{2})^{n}}$$

by the n-th root test.

Wolfram Alpha gives me a step by step solution that involves L'Hopital's Rule but Im not allowed to use it yet. Any ideas?

Thanks :)

Answer 1

Using the root test doesn't seem the best way: the ratio test requires $$ \lim_{n\to\infty}\frac{3n+2}{(\sqrt{2})^{n+1}}\frac{(\sqrt{2})^n}{3n-1} =\frac{1}{\sqrt{2}}<1 $$

For the root test: you need $$ \lim_{n\to\infty}\frac{(3n-1)^{1/n}}{\sqrt{2}} $$ So you just need the limit of the numerator and taking the logarithm brings you to $$ \lim_{n\to\infty}\frac{\log(3n-1)}{n}=0 $$ Hence $\lim_{n\to\infty}(3n-1)^{1/n}=e^0=1$.

Why is that limit zero? Compute instead $$ \lim_{x\to\infty}\frac{\log(3x-1)}{x}= \lim_{x\to\infty}\frac{\log(3x-1)}{3x-1}\frac{3x-1}{x} $$ The second factor has limit $3$. For the first do the change of variables $3x-1=e^t$, so you get $$ \lim_{t\to\infty}\frac{t}{e^t} $$ which you should know is zero.

Answer 2

By noticing that $$\sqrt[3n-1]{3n-1} \le \sqrt[n]{3n-1} \le \sqrt[n]{3n} = \sqrt[n]3 \sqrt[n]n$$ and using the fact that $\lim\limits_{n\to\infty} \sqrt[n]n=1$; see here or here we get that $$\lim\limits_{n\to\infty} \sqrt[n]{3n-1} = 1.$$

Hence $$\lim\limits_{n\to\infty} \frac{\sqrt[n]{3n-1}}{\sqrt2} = \frac1{\sqrt2}.$$