Finding the Limit Inferior and Limit Superior of a Sequence

[evinda]

Hello! (Wave)

I want to compute liminf and limsup of $\left( (-1)^{n^3} \left( 1+\frac{1}{n}\right)^n\right)$.

I have thought the following so far:

From definition we have that $\lim \inf x_n=\lim_{n \to \infty} \left( inf_{k \geq n} x_k \right)$ and $\lim \sup x_n=\lim_{n \to \infty} \left( \sup_{k \geq n} x_k \right)$.

If $k$ is odd, then $(-1)^{k^3} \left( 1+\frac{1}{k}\right)^k=-\left( 1+\frac{1}{k}\right)^k$.

If $k$ is even , then $(-1)^{k^3} \left( 1+\frac{1}{k}\right)^k=\left( 1+\frac{1}{k}\right)^k$.

It holds that $-\left( 1+\frac{1}{k}\right)^k \geq - \left( 1+\frac{1}{n}\right)^k$ and $\left( 1+\frac{1}{k}\right)^k \leq \left( 1+\frac{1}{n}\right)^k$.But we cannot bound $\left( 1+\frac{1}{n}\right)^k$ and $- \left( 1+\frac{1}{n}\right)^k$ by an expression of $n$, can we? (Thinking)

If not, how can we compute liminf and limsup?
.

 

[Olinguito]

Hi evinda.

Are you aware of that $\displaystyle \lim_{n\to\infty}\left(1+\frac1n\right)^n=e$?

 

[evinda]

Hi evinda.

Are you aware of that $\displaystyle \lim_{n\to\infty}\left(1+\frac1n\right)^n=e$?

Yes, but how can we use this in our case? (Thinking)

 

[I like Serena]

Hey evinda!

Suppose we can find an upper bound, such that the sequence gets arbitrarily close to it.
Wouldn't that qualify as a $\limsup$? (Wondering)

Suppose we pick a subsequence such that all elements are positive, can we find it's limit?

 

[evinda]

Hey evinda!

Suppose we can find an upper bound, such that the sequence gets arbitrarily close to it.
Wouldn't that qualify as a $\limsup$? (Wondering)

Suppose we pick a subsequence such that all elements are positive, can we find it's limit?

We get such a subsequence by picking only the even terms, its limit will be $e$.

But will this also be the limsup ? If so, why?

 

[I like Serena]

We get such a subsequence by picking only the even terms, its limit will be $e$.

But will this also be the limsup ? If so, why?

All elements in the sequence are smaller than $e$, so $e$ is an upper bound.
Any value that is smaller than $e$ is not an upper bound, since there is a sub sequence (the even $n$) with $e$ as its limit.
Therefore $e$ is the supremum of the sequence ($\sup x_k = e$).
If we start the sequence at $n$, we still have a sub sequence (the even $n$) that has $e$ as its limit.
So for any $n$:
$$\sup_{k\ge n} x_k = e$$
Thus:
$$\limsup_{n\to\infty} x_n = \lim_{\vphantom{\large k}n\to\infty}(\sup_{k\ge n} x_k) = \lim_{n\to\infty} e = e$$
(Thinking)

 

[evinda]

All elements in the sequence are smaller than $e$, so $e$ is an upper bound.
Any value that is smaller than $e$ is not an upper bound, since there is a sub sequence (the even $n$) with $e$ as its limit.
Therefore $e$ is the supremum of the sequence ($\sup x_k = e$).
If we start the sequence at $n$, we still have a sub sequence (the even $n$) that has $e$ as its limit.
So for any $n$:
$$\sup_{k\ge n} x_k = e$$
Thus:
$$\limsup_{n\to\infty} x_n = \lim_{\vphantom{\large k}n\to\infty}(\sup_{k\ge n} x_k) = \lim_{n\to\infty} e = e$$
(Thinking)

I see... thanks a lot! (Happy)