# Compute liminf, limsup

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?

• Whenever you see an expression of the form $(1+{x \over n})^n$ you should be thinking $e^x$. Commented Dec 1, 2018 at 21:58

We have that

$$\left| (-1)^{n^3} \left( 1+\frac{1}{n}\right)^n\right|\le e$$

and

• $$n^3$$ even $$\implies (-1)^{n^3}\left( 1+\frac{1}{n}\right)^n \to e$$

• $$n^3$$ odd $$\implies (-1)^{n^3}\left( 1+\frac{1}{n}\right)^n \to -e$$

then we can conclude that $$\limsup=e$$ and $$\liminf=-e$$.

• Simple and on point as always ! +1 Commented Dec 1, 2018 at 21:55
• @Rebellos Thanks, much appreciative! Bye
– user
Commented Dec 1, 2018 at 21:56
• Why does it hold that $\left| (-1)^{n^3} \left( 1+\frac{1}{n}\right)^n\right| \leq e$ ? Commented Dec 1, 2018 at 22:03
• @Evinda If we can show that $|a_n| \le L$ and we find a subsequence such that $a_{n_k} \to L$ therefore $\limsup a_n=L$. And similarly for $\liminf$. It is a theorem.
– user
Commented Dec 1, 2018 at 22:06
• @Evinda We have that $$\left| (-1)^{n^3} \left( 1+\frac{1}{n}\right)^n\right| =\left( 1+\frac{1}{n}\right)^n\leq e$$
– user
Commented Dec 1, 2018 at 22:06