Better way to do $\lim_{n\to\infty}\frac{(3n-1)^{1/n}}{{\sqrt{2}}^{1/n}}$ 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 :)
 A: 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.
A: 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
$$\lim\limits_{n\to\infty} \frac{\sqrt[n]{3n-1}}{\sqrt2} = \frac1{\sqrt2}.$$
