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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

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    $\begingroup$ Since $n/(2n-1)\to1/2$ as $n\to\infty$, there exists $N$ with $n/(2n-1)<3/4$ for all $n>N$. $\endgroup$ Jun 21 '18 at 17:47
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Observe that $n \ge 2$ is equivalent to $\frac{n}{2n-1} \le \frac{2}{3}$. So, for $n\ge 2$ we have $\left(\frac{n}{2n-1}\right)^n\le \left(\frac{2}{3}\right)^n$

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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 $n \ge n_0:$

$b^n \lt \dfrac{1}{nx} \le \dfrac{1}{n_0x} \lt\dfrac{1}{Mx} = \epsilon.$

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Hint:

By continuity, determine only the limit of the log and use equivalents:

$\dfrac n{2n-1}\sim_\infty\dfrac 12$, so $$\log\Bigl(\frac n{2n-1}\Bigr)^n=n\log\Bigl(\frac n{2n-1}\Bigr)\sim_\infty n\log\frac12=-n\log 2\to-\infty\qquad\text{as }\; n\to +\infty.$$

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  • $\begingroup$ Thank you, this is a nice solution. Could I also use the same argument, but instead of the logarithm I used $f(x)=\sqrt[n](x)$ since it is continuous? $\endgroup$ Jun 21 '18 at 17:55
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    $\begingroup$ It's more complex to do it directly, because the definition of $\sqrt[n]{\;}$ is through an exponential, and you can't compose equivalents by an exponential as easily as with logarithm. Roughly speaking, $f\sim g \not \Rightarrow \mathrm e^f\sim \mathrm e^g$ in all cases. $\endgroup$
    – Bernard
    Jun 21 '18 at 18:00
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We have that

$$\bigg(\dfrac{n}{2n-1}\bigg)^n=\left(\frac12\right)^n\bigg(\dfrac{2n}{2n-1}\bigg)^n=\left(\frac12\right)^n\left[\bigg(1+\dfrac{1}{2n-1}\bigg)^{2n-1}\right]^{\frac{n}{2n-1}}\to 0\cdot\sqrt e=0$$

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Consider $b_n = a_n ^ {1/n}$ and show that $b_n$ converges to $1/2 < 1 $.

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    $\begingroup$ Instead of \$b_n\$ = \$a_n\$ \$^\$ \$1/n\$ try \$b_n=a_n^{1/n}\$ instead, and in general it's way better to avoid multiply dollar signs whenever possible. $\endgroup$
    – kingW3
    Jun 21 '18 at 18:39
  • $\begingroup$ Thanks a lot man!! $\endgroup$
    – al.al.
    Jun 21 '18 at 19:09

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