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Let $$x_n=(-1)^n \left(2+\frac{3^n}{n!}+\frac{4}{n^2}\right)$$ and find the upper and lower limits of the sequence $\{x_n\}_{n=1}^\infty$.

We put $n=1, 2, 3 \dots $ then $x_1=-9, x_2 = \frac{15}{2}, \dots $ after some stage we see that limit superior is $x_2$ and inferior is $x_1$ is it right? Please explain.

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    $\begingroup$ do $x_{2n}$ and $x_{2n+1}$ $\endgroup$ – janmarqz Jul 17 '18 at 3:46
  • $\begingroup$ The limit superior of a sequence, is not the largest term of the sequence. Please see the definition of limit superior again : it is the largest limit point of the sequence as a set, or the largest number such that there is a subsequence of the given sequence converging to that point. $\endgroup$ – астон вілла олоф мэллбэрг Jul 17 '18 at 4:04
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Notice

$\underset{n \to \infty}{\lim \sup}\, x_n = \inf \{x_{2n} : n \in \mathbb{N}\} $,

and

$\underset{n \to \infty}{\lim \inf}\, x_n = \sup \{x_{2n-1} : n \in \mathbb{N}\} $.

Moreover, $2$ is a lower bound for the set $\left\{2+\frac{3^{2n}}{(2n)!}+\frac{1}{n^2} : n \in \mathbb{N}\right\}$, and $-2$ is an upper bound for the set $\left\{-2-\frac{3^{2n-1}}{(2n-1)!}-\frac{4}{(2n-1)^2} : n \in \mathbb{N}\right\} $.

Also observe that we have $\frac{3^n}{n!} < 2^{6-n}$ whenever $n>6$.

Let $\varepsilon>0$ be given. By the Archimedean Property we may find positive integers $N_1$ and $N_2$ such that \begin{aligned}&N_1 > 7 - \frac{\log\varepsilon}{\log 2}, \text{ and} \\& N_2\varepsilon > 8 . \end{aligned} Set $N=\max\{7, N_1, N_2\}$. So if $n \geq N$, then we have \begin{equation}\frac{3^n}{n!}+\frac{4}{n^2}<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon . \end{equation} So there is $N \in \mathbb{N}$ such that \begin{aligned} &x_{2N}<2+\varepsilon, \text{ and} \\& x_{2N-1}>-2-\varepsilon. \end{aligned} Therefore, $\underset{n \to \infty}{\lim \sup}\, x_n =2$ and $\underset{n \to \infty}{\lim \inf}\, x_n =-2$.

Note: Since $N \geq 7$, we know that $2N-1>N$ and $2N>N$.

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  • $\begingroup$ Is there a reason this got down voted? Because I would love to hear it. $\endgroup$ – Matt A Pelto Jul 17 '18 at 5:02
  • $\begingroup$ I guess it is because I accidentally had written "$\underset{n \to \infty}{\lim \sup}\, = \inf \{x_{2n} : n \in \mathbb{N}\} $" and "$\underset{n \to \infty}{\lim \inf} = \sup \{x_{2n-1} : n \in \mathbb{N} \}$", where obviously I did not notice that I forgot to write "$x_n$". I really wish I knew what sort of person down votes for this. They could have easily edited my answer or left a comment...clearly a lazy ingrate. $\endgroup$ – Matt A Pelto Jul 17 '18 at 5:15
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Here is a less rigorous but quicker way to look at the problem.

We have $$x_n=(-1)^n \left(2+\frac{3^n}{n!}+\frac{4}{n^2}\right)$$

Let's look at $$\lvert x_n\rvert=2+\frac{3^n}{n!}+\frac{4}{n^2}$$

We know that $\dfrac4{n^2}$ is decreasing and $2$ is constant.

The first few terms of $\dfrac{3^n}{n!}$ are $$\dfrac{3}{1},\, \dfrac{3\cdot3}{2\cdot1},\, \dfrac{3\cdot3\cdot3}{3\cdot2\cdot1},\, \dfrac{3\cdot3\cdot3\cdot3}{4\cdot3\cdot2\cdot1},\, \dfrac{3\cdot3\cdot3\cdot3\cdot3}{5\cdot4\cdot3\cdot2\cdot1},\cdots$$

which has 2nd and 3rd terms equal and 4th terms onwards decreasing.

All in all, $\lvert x_n\rvert$ will be decreasing from 3rd terms onwards. Hence, $x_1$ and $x_2$ are the lower and upper limits respectively.

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