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

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?

• $\frac{n}{10} > n?$ – Sahiba Arora Aug 15 '17 at 17:06
• Perhaps better to say $\frac{n}{10} > N > B$ when $n > 10N$, in which case $\left(\frac{11}{10}\right)^n > B$ – Henry Aug 15 '17 at 17:10
• Please replace the incorrect Show that lim x_n is divergent by Show that (x_n) is divergent or, but this is not as good, Show that lim x_n = +oo. – Did Aug 15 '17 at 17:21
• I don't see the point of the separate N and B, and saying "let epsilon > 0" seems misplaced - you are really just invoking Bernoulli's inequality, and you just want to say it holds for all epsilon and n. (If you first say "let epsilon > 0" then you should not choose a specific epsilon afterwards.) – TMM Aug 15 '17 at 17:27
• @SahibaArora Hah! Yes!. Sorry. – James Ensor Aug 15 '17 at 17:29

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

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.

• This is in fact the essence of the OP's almost correct proof, cleaned up so that it can be expressed in one or two lines. +1 for the nice editing. – Ethan Bolker Aug 15 '17 at 17:41

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.