Bernoulli's inequality

[william_33]

Let \(\displaystyle d\in\mathbb{R}\) satisfy \(\displaystyle d>1\). Use Bernoulli's inequality and by the definition of a limit to show that the sequence \(\displaystyle d^n\) is not bounded in \(\displaystyle \mathbb{R}\), hence not convergent.

My attempt:

Let where

We know from Bernoulli's inequality that \(\displaystyle (1+a)^n\ge (1+na)\) hence since \(\displaystyle d>1\) then \(\displaystyle 1<d^n=\frac{1}{(1+a)^n}\le\frac{1}{1+na}\le\frac{1}{na}\), but I get stuck here because I do not know how to prove by the definition of a limit that it is in fact divergent.

 

[pka]

Let \(\displaystyle d\in\mathbb{R}\) satisfy \(\displaystyle d>1\). Use Bernoulli's inequality and by the definition of a limit to show that the sequence \(\displaystyle d^n\) is not bounded in \(\displaystyle \mathbb{R}\), hence not convergent.

Note that \(\displaystyle \exists a>0\) such that \(\displaystyle d=1+a\).
So \(\displaystyle d^n=(1+a)^n\ge 1+na\).

Also \(\displaystyle \forall N\in\mathbb{Z}^+,~\exists k[k\cdot a\ge N\). That means \(\displaystyle d^k\ge 1+N\).

Can the sequence \(\displaystyle (d^n)\) be bounded?

Recall that convergent sequences are bounded.

 

[william_33]

"Also, \(\displaystyle \forall N\in \mathbb{Z}^+\), there exists \(\displaystyle k[k\dot\ a\ge N\), That means \(\displaystyle d^k\ge 1 + N\)."

What do you mean by this notation \(\displaystyle k[k\dot\ a\ge N\)? Also, how did you get that \(\displaystyle d^k\ge 1 + N\)? I know that this sequence cannot be bounded since it is not convergent.

 

[daon2]

It means that for any \(\displaystyle N>0\) the monotone sequence \(\displaystyle \{1+na\}\) "eventually" surpasses \(\displaystyle 1+N\). Since \(\displaystyle d^n \ge 1+na\) for all \(\displaystyle n\), this shows divergence.

Edit: Also, "not convergent" does not imply "not bounded."

 

[william_33]

Thanks Daon!