Prove that $\lim_{n\to \infty} (1+\frac{1}{n^3})^n = 1$ using definition of limit

Question

I cannot seem to understand how to prove $\lim_{n\to \infty} (1+\frac{1}{n^3})^n = 1$ through the usual $|a_n-l|<\epsilon$. I must do this using no other theorem, just the definition.

Doing it for $\lim_{n\to \infty} (1+\frac{1}{n^3}) = 1$ is really easy, but I do not know how to handle the exponentiation.

Thank you very much for any help.

EDIT: An example of proof would go like this:

$\epsilon>0, \forall N>n$

$|1-\frac{1}{n^3}-1|=|-\frac{1}{n^3}|=\frac{1}{n^3}<\frac{1}{n}<\frac{1}{N}<\epsilon$

$N>\frac{1}{\epsilon}$

Clarification:

$\epsilon>0, \forall N>n$

I need to find what is the value of $N$ in relation to $\epsilon$ when:

$|(1-\frac{1}{n^3})^n-1|<\epsilon$ (the only manipulation allowed being the ones I showcased in my example).

Answer 1

By Bernoulli's inequality for $n\geq 2$ $$\left(1-\frac1{n^3+1}\right)^n\geq 1-\frac{n}{n^3+1}>1-\frac1{n^2}$$ and taking reciprocals $$\left(1+\frac1{n^3}\right)^n < \frac1{1-\frac1{n^2}} = 1+\frac1{n^2-1}$$

Answer 2

Let $$a_n=(1+\frac{1}{n^3})^n$$

Given an $\epsilon>0$,

We have

$$ln(a_n)=n ln(1+\frac{1}{n^3})$$

$$=n(\frac{1}{n^3}(1+\epsilon(n))),$$

with $\lim_{n\to\infty}\epsilon(n)=0$.

this gives

$$ln(a_n)=\frac{1}{n^2}(1+\epsilon(n))$$

and

$$a_n=e^{\frac{1}{n^2}(1+\epsilon(n)) }$$

or

$$a_n=1+\frac{1}{n^2}(1+\epsilon(n))$$

which implies

$$|a_n-1|=\frac{1}{n^2}(1+\epsilon(n)).$$

$\exists N_1 \in \mathbb N $ such that

$$n\geq N_1 \implies -1<\epsilon(n)<1$$

and $N_2$ such that $n\geq N_2 \implies \frac{2}{n^2}<\epsilon$ given by

$N_2=\lfloor \sqrt{\frac{2}{\epsilon}}\rfloor +1$.

at the end, we take $N=max(N_1,N_2)$

we check that

$$n\geq N \implies |a_n-1|<2\frac{\epsilon}{2}.$$

Answer 3

Elaborating on lisyarus comment:

$$ \left(1+\frac{1}{n^3}\right)^n=\sum_{k=0}^n {{n}\choose{k}}\frac{1}{n^{3k}} $$ and therefore

$$ \left|\left(1+\frac{1}{n^3}\right)^n-1\right|=\sum_{k=1}^n {{n}\choose{k}}\frac{1}{n^{3k}}=R(n) $$

Now, you need to find $N$ such that $n>N$ implies $R(n)<\varepsilon$. This can be done using the fact that

$$ {{n}\choose{k}}=\frac{n(n-1)\cdots(n-k+1)}{k!}\leq \frac{n^k}{k!}. $$

But then

$$ R(n)\leq \sum_{k=1}^{n}\frac{n^k}{k!}\frac{1}{n^{3k}}=\sum_{k=1}^{n}\frac{1}{k!n^{2k}}\leq \sum_{k=1}^{n}\frac{1}{n^2}=\frac{1}{n}. $$

Therefore, if one takes $N=1/\varepsilon$, we have that $n>N$ implies $R(n)\leq\varepsilon$ (strictly if $\varepsilon<1$).