Proving limits of two sequences I've trouble proving that:
a) $\lim(\frac{1}{\sqrt[n]{2}-1} - \frac{2}{\sqrt[n]{4}-1})=\frac{1}{2}$
b) $\lim \frac{2n-1}{2n+1}\cdot  \frac{2n-2}{2n+2}\cdot ... \cdot \frac{n}{3n}=0$.
In a) I've tried using  formula $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+...+b^{n-1})$ which gives me $\lim(\frac{1}{\sqrt[n]{2}-1} - \frac{2}{\sqrt[n]{4}-1})=\lim \sum_{k=1}^{n}\frac{2}{\sqrt[n]{2^k}}(1- \frac{4}{3\sqrt[n]{2^k}})$ and I can't go any further.
In b) I've tried using Squeeze Theorem $\lim \frac{2n-1}{2n+1}\cdot  \frac{2n-2}{2n+2}\cdot ... \cdot \frac{n}{3n}>(\frac{1}{3})^n$ but i cant find second inequality.
 A: *

*One may recall that, by the Taylor series expansion, as $n \to
    \infty$, $$ \sqrt[n]{2}=e^{\large\frac{\ln 2}{n}}=1+\frac{\ln
    2}{n}+\frac{\ln^2 2}{2n^2}+\mathcal{O}\left(\frac1{n^3}\right) $$
giving $$ \frac{1}{\sqrt[n]{2}-1}=\frac{n}{\ln
2}-\frac12+\mathcal{O}\left(\frac1{n}\right) $$ similarly $$
\frac{2}{\sqrt[n]{4}-1}=\frac{n}{\ln
2}-1+\mathcal{O}\left(\frac1{n}\right) $$ then, as $n \to \infty$,
$$
\frac{1}{\sqrt[n]{2}-1}-\frac{2}{\sqrt[n]{4}-1}=\frac12+\mathcal{O}\left(\frac1{n}\right)
$$ and one gets the announced limit.

*One may observe that, as $n \to \infty$, $$ \frac{2n-1}{2n+1}\cdot 
\frac{2n-2}{2n+2} \cdots  \frac{n}{3n}=\frac12\cdot\frac{(2n)!}{n!} \cdot
\frac{(2n)!}{(3n)!} \sim \left(\frac{16}{27} \right)^n\cdot \frac{2
n}{\sqrt{3}} \to 0$$ by using Stirling's formula.

A: For part a), we have
$${1\over2^{1/n}-1}-{2\over4^{1/n}-1}={1\over2^{1/n}-1}-{2\over2^{2/n}-1}={1\over2^{1/n}-1}-{2\over(2^{1/n}-1)(2^{1/n}+1)}={(2^{1/n}+1)-2\over(2^{1/n}-1)(2^{1/n}+1)}={1\over2^{1/n}+1}\to{1\over1+1}={1\over2}$$
For part b), we have
$$\left(2n-1\over2n+1\right)\left(2n-2\over2n+2\right)\cdots\left(n+1\over3n-1\right)\left(n\over3n\right)=\left(n\over2n+1\right)\left(n+1\over2n+2\right)\cdots\left(2n-2\over3n-1\right)\left(2n-1\over3n\right)$$
and, for $0\le k\le n-1$, we have 
$${n+k\over2n+1+k}\lt{2\over3}$$
since $3n+3k\lt4n+2+2k$ is equivalent to $k\lt n+2$.
