# Thread: Limit points, limsup και liminf

1. Hey!!

I want to find the $\lim\sup$, $\lim\inf$ and the limit points of the following sequences:
1. $a_n=(-1)^n\frac{3n+4}{n+1}$
2. $a_n=\sqrt[n]{n+(-1)^nn}$
3. $a_n=\left ( \frac{n+(-1)^n}{n}\right )^n$
4. $a_n=(-1)^{\frac{n(n+1)}{2}}\sqrt[n]{1+\frac{1}{n}}$

I have done the following:

1. $$a_{2k}=(-1)^{2k}\frac{3(2k)+4}{2k+1}=\frac{6k+4}{2k+1}\Rightarrow \lim_{k\rightarrow \infty}a_{2k}=3 \\ a_{2k-1}=(-1)^{2k-1}\frac{3(2k-1)+4}{2k-1+1}=-\frac{6k+1}{2k}\Rightarrow \lim_{k\rightarrow \infty}a_{2k-1}=-3$$
So, $(a_n)$ has the limit points $3$ and $-3$.
Therefore, $\lim\sup a_n=3$ and $\lim\inf a_n=-3$.

2. $$a_{2k}=\sqrt[2k]{2k+(-1)^{2k}2k}=\sqrt[2k]{4k}=2^{\frac{1}{k}}\left ( k^{\frac{1}{k}}\right )^{\frac{1}{2}}\Rightarrow \lim_{k\rightarrow \infty}a_{2k}=1\\ a_{2k-1}=\sqrt[2k-1]{2k-1+(-1)^{2k-1}(2k-1)}=0\Rightarrow \lim_{k\rightarrow \infty}a_{2k-1}=0$$
So, $(a_n)$ has the limit points $1$ and $0$.
Therefore, $\lim\sup a_n=1$ and $\lim\inf a_n=0$.

3. $$a_{2k}=\left ( \frac{2k+(-1)^{2k}}{2k}\right )^{2k}=\left (\frac{2k+1}{2k}\right )^{2k}=\left (1+\frac{1}{2k}\right )^{2k}\Rightarrow \lim_{k\rightarrow \infty}a_{2k}=e \\ a_{2k-1}=\left ( \frac{2k-1+(-1)^{2k-1}}{2k-1}\right )^{2k-1}=\left (\frac{2k-1-1}{2k-1}\right )^{2k-1}=\left (1-\frac{1}{2k-1}\right )^{2k-1}\Rightarrow \lim_{k\rightarrow \infty}a_{2k-1}=e^{-1}$$
So, $(a_n)$ has the limit points $e$ and $e^{-1}$.
Therefore, $\lim\sup a_n=e$ and $\lim\inf a_n=e^{-1}$.

Is everything correct so far? How could we prove that the limit points that I found at each case are the only ones?

4. $$a_n=(-1)^{\frac{n(n+1)}{2}}\sqrt[n]{1+\frac{1}{n}}$$
I tried to find again the limits of $a_{2k}$ and $a_{2k-1}$ but I failed.
Do we find the limit points in this case in an other way?

2. Originally Posted by mathmari
Is everything correct so far? How could we prove that the limit points that I found at each case are the only ones?

4. $$a_n=(-1)^{\frac{n(n+1)}{2}}\sqrt[n]{1+\frac{1}{n}}$$
I tried to find again the limits of $a_{2k}$ and $a_{2k-1}$ but I failed.
Do we find the limit points in this case in an other way?
Hey mathmari!!

They look good to me.

How about trying $a_{4k},\ a_{4k+1},\ a_{4k+2}$, and $a_{4k+3}$?

Originally Posted by I like Serena
They look good to me.

Originally Posted by I like Serena
How about trying $a_{4k},\ a_{4k+1},\ a_{4k+2}$, and $a_{4k+3}$?
Calculating these we get that the limit points are $1$ and $-1$, so $\lim\sup a_n=1$ and $\lim\inf a_n=-1$, right?

Instead of these we could also calculate the subsequences $a_{4k},\ a_{4k-1},\ a_{4k-2}$, and $a_{4k-3}$, or not?

4. Yes και yes.