# Example of $\liminf_{n\rightarrow\infty}(x_n+y_n)>\liminf_{n\rightarrow\infty}x_n+\liminf_{n\rightarrow\infty}y_n$

A result from introductory analysis shows that given two bounded sequences $\{x_n\}$ and $\{y_n\}$,

$\liminf_{n\rightarrow\infty}(x_n+y_n)\geq\liminf_{n\rightarrow\infty}x_n+\liminf_{n\rightarrow\infty}y_n$.

I need to find at least two examples of where the above inequality is strict. One is if $x_n=(-1)^n$ and $y_n=(-1)^{n-1}$. Can anyone think of any other examples? I need at least one more.

• $x_n = \sin(n)$ and $y_n = \cos(n)$?
– M.B.
Apr 17, 2014 at 1:23
• I don't think those sequences have any cluster points. If you meant $\{\sin(\frac{\pi}{2}-n\pi)\}$ and $\{\cos(n\pi)\}$, they're the same sequences, just written differently.
– mjh
Apr 17, 2014 at 1:26
• @mjh: no, that sequence is good: the smallest that $\sin(n)+\cos(n)=\sqrt2\sin(n+\pi/4)$ can be is $-\sqrt2$, yet each alone can reach $-1$.
– robjohn
Apr 17, 2014 at 1:39

I might as well give my comment in an answer. Try the following:

$$x_n = \sin(n)\qquad y_n = \cos(n)$$

For all $x$ we have that $\sin(x) + \cos(x) \geq -\sqrt{2}$, and $\sin(x) \geq -1$, $\cos(x) \geq -1$.

• Yes, but what are the limit inferiors of $\{x_n\}$, $\{y_n\}$, and $\{x_n+y_n\}$?
– mjh
Apr 17, 2014 at 1:31
• $-1, -1, -\sqrt{2}$.
– M.B.
Apr 17, 2014 at 1:34
• The limit inferior of a bounded sequence with cluster points is the smallest cluster point of the sequence. What you're saying is that the sequence $\{-1\}$ is a subsequence of both $\{\sin(n)\}$ and $\{\cos(n)\}$. But $\sin(n)=-1$ or $\cos(n)=-1$ only if $n$ involves $\pi$, and $n$ is a natural number.
– mjh
Apr 17, 2014 at 1:40
• @mjh: $\sin(n)+\cos(n)=\sqrt2\sin(n+\pi/4)$
– robjohn
Apr 17, 2014 at 1:40
• @mjh: sure but, $1,2,3,\ldots \pmod \pi$ comes arbitrarily close
– M.B.
Apr 17, 2014 at 1:46