Show that $\bigg(\dfrac{n}{2n-1}\bigg)^n$ goes to 0 I have the sequence defined:
$$\bigg(\dfrac{n}{2n-1}\bigg)^n$$
and I have to show that it goes to zero. For a similar sequence, in order to show that it converges to 0, I would show:
\begin{align}
&\bigg|\dfrac{n}{2n+1}\bigg|^n\\
<&\bigg(\dfrac{n}{2n}\bigg)^n\\
=&\bigg(\dfrac{1}{2}\bigg)^n\\
<&\epsilon\\
\end{align}
Now in the case of the sequence defined in the title, can I "reduce" the fraction in a neat way as above?
My attempt was as follows:
\begin{align}
&\bigg|\dfrac{n}{2n-1}\bigg|^n\\
<&\bigg|\dfrac{1}{2n-1}\bigg|^n\\
=&\dfrac{1}{(2n-1)^n}\\
\end{align}
Now the last term is obviously $<\epsilon \ \exists N \ \forall n>N$. However I am not managing to show this due to having an $n$ in the exponent. How shall I go about this? Thanks in advance
 A: Observe that $n \ge 2$ is equivalent to $\frac{n}{2n-1} \le \frac{2}{3}$. So, for $n\ge 2$ we have $\left(\frac{n}{2n-1}\right)^n\le \left(\frac{2}{3}\right)^n$
A: Let $n \gt 2$:
$(\dfrac{n}{2n-1})^n$ =$(\dfrac{1}{2-1/n})^n \lt$
$(\dfrac{1}{2-1/2})^n = (\dfrac{2}{3})^n.$
Need to show that for $0<b<1$ , 
$\lim_{n \rightarrow \infty} b^n=0.$
$b=\dfrac{1}{1+x}$ , $x >0.$
$b^n =\dfrac{1}{(1+x)^n} \lt $
$\dfrac{1}{1+nx} \lt \dfrac{1}{nx}.$
Let $\epsilon >0$ be given:
$ M:= \dfrac{1}{x\epsilon}$.
Archimedean Principle:
There is a $n_0$, $n_0 \in \mathbb{Z^+}$, 
such that $n_0 >M.$
Then for $n \ge n_0:$
$b^n \lt \dfrac{1}{nx} \le \dfrac{1}{n_0x} \lt\dfrac{1}{Mx} = 
\epsilon.$
A: Consider $b_n = a_n ^ {1/n}$ and show that $b_n$ converges to $1/2 < 1 $. 
A: Hint:
By continuity, determine only the limit of the log and use equivalents:
$\dfrac n{2n-1}\sim_\infty\dfrac 12$, so 
$$\log\Bigl(\frac n{2n-1}\Bigr)^n=n\log\Bigl(\frac n{2n-1}\Bigr)\sim_\infty n\log\frac12=-n\log 2\to-\infty\qquad\text{as }\; n\to +\infty.$$
A: We have that
$$\bigg(\dfrac{n}{2n-1}\bigg)^n=\left(\frac12\right)^n\bigg(\dfrac{2n}{2n-1}\bigg)^n=\left(\frac12\right)^n\left[\bigg(1+\dfrac{1}{2n-1}\bigg)^{2n-1}\right]^{\frac{n}{2n-1}}\to 0\cdot\sqrt e=0$$
