Consider the sequence defined by $u_n=(\frac{2n}{3n+1})^n$, $n \in \Bbb{Z^+}$.

(a) Show that $\frac{1}{2} \le \frac{2n}{3n+1} < \frac{2}{3}$

(b) Hence use the squeeze theorem to find $\lim \limits_{n \to \infty} u_n$.

For (a) I started as follows:

$\frac{1}{2} < \frac{2}{3} \le \frac{2}{3}$

$\frac{1}{2} < \frac{2n}{3n} \le \frac{2}{3}$

But then I'm now really sure how to add the $1$ to the numerator.

I tried working backwards i.e. by simplifying the expression into:

$\frac{1}{2} < \frac{2n}{3n+1} < \frac{2}{3}$ $\implies$ $3<3n<3n+4$. This seems to be true for $n\ge 2$, so technically this would be prove that I can apply the squeeze theorem, right? But I don't think the authors of the book expected me to take this approach. What is the best way to do part (a)?

Part (b) confuses me since even if I did part (a), for the squeeze theorem to apply the sequences of the left and right have to have the same limit but in this case the limits would be different ($\frac{1}{3}$ and $\frac{2}{3}$, respectively). How do I handle that?

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    $\begingroup$ For the only inequality we really need, $\frac{2n}{3n+1}\lt \frac{2n}{3n}=\frac{2}{3}$. The limit is $0$, for $(2/3)^n\to 0$ as $n\to\infty$. $\endgroup$ – André Nicolas Mar 29 '16 at 20:59
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    $\begingroup$ Regarding part (b), you say that the limits would be $\frac{1}{3}$ and $\frac{2}{3}$ respectively. But you are forgetting to take $n^{\text{th}}$ powers. $\endgroup$ – Lee Mosher Mar 29 '16 at 21:04

One may observe that, for $n=1,2,3,\ldots$, one has

  • $$ \frac{2n}{3n+1} - \frac{2}{3}=-\frac{2}{3+9 n}<0 $$

  • $$ \frac{2n}{3n+1} - \frac12=\frac{n-1}{2+6 n}\geq0 $$

giving $$ \frac12\leq \frac{2n}{3n+1} < \frac{2}{3} $$ and, since $x \mapsto x^n$ is increasing, $$ \left(\frac12\right)^n\leq \left(\frac{2n}{3n+1}\right)^n < \left(\frac23\right)^n $$ then one may conclude using $$ \lim_{n \to \infty}\left(\frac12\right)^n=\lim_{n \to \infty}\left(\frac23\right)^n=0. $$


Your approach to (a) is correct, and indeed you need $n\geq 2$ for the inequality to hold.

For part (b) , take $n$th powers of the inequality (valid because all terms are positive) to get

$$ \left(\frac12\right)^n < u_n < \left(\frac{2}{3}\right)^n. $$

Now $u_n$ is between two sequences that both tend to zero, so you can apply the squeeze theorem.


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