Show that $\left( \frac {11} {10}\right) ^{n}$ is divergent.

Question

Show that $\left( \dfrac {11} {10}\right) ^{n}$ is divergent.

My proof. Let $B\in\mathbb{R}$. By the Archimedean property there is a $N$ in $\mathbb{N}$ such that $N>B$.
Let $\varepsilon >0$ By the Bernoulli inequality, we have $\left( 1+\varepsilon \right) ^{n}\geq 1+n\varepsilon$ for all $n\in\mathbb{N}$. Now, take $\varepsilon=( \dfrac {11} {10}-1)$. Then, we obtain, $\left( \dfrac {11} {10}\right) ^{n}\geq \dfrac {n} {10}+1$. So, for all $n\geq N$ we have $\dfrac {n} {10}+1>\dfrac {n} {10}>n>N>B.$ Thus, since $\left( \dfrac {11} {10}\right) ^{n}\geq \dfrac {n} {10}+1$, $\left( \dfrac {11} {10}\right) ^{n}>B$
for all $n\geq N$.

We are done.

Can you check my proof?

Answer 1

The error in your proof is the assertion $$\frac{n}{10}>n$$ This is not true as $n \in \mathbb{N}.$

In the comments, Henry suggests how your proof can be fixed.

An alternate way to prove the sequence is divergent is to show that it is unbounded.

It follows by Bernoulli's inequality that $$\left(\frac{11}{10}\right)^n=\left(1+\frac{1}{10}\right)^n\geq1+\frac{n}{10}>\frac{n}{10}$$ for all $n \in \mathbb N.$ Hence, the sequence is unbounded.

Answer 2

Easy to think solution:

Note that $\ln$ is increasing function.

Note that $\ln\Big(\dfrac{11}{10}\Big)=\ln11-\ln10=c>0$

Now $\ln\Big(\dfrac{11}{10}\Big)^n=n(\ln11-\ln10)=nc$

Now since $c>0$, for every $N\in \mathbb{N}$ and $N>\Big\lfloor\dfrac{1}{c}\Big\rfloor+1$, you can find a $n\in\mathbb{N}$ such that $nc>N$. Hence $\ln\Big(\dfrac{11}{10}\Big)^n$ diverges to infinity. Since $\ln$ is increasing function $\Big(\dfrac{11}{10}\Big)^n$ also diverges to infinity.