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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?

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    $\begingroup$ $\frac{n}{10} > n?$ $\endgroup$ – Sahiba Arora Aug 15 '17 at 17:06
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    $\begingroup$ Perhaps better to say $\frac{n}{10} > N > B$ when $n > 10N$, in which case $\left(\frac{11}{10}\right)^n > B$ $\endgroup$ – Henry Aug 15 '17 at 17:10
  • $\begingroup$ 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. $\endgroup$ – Did Aug 15 '17 at 17:21
  • $\begingroup$ 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.) $\endgroup$ – TMM Aug 15 '17 at 17:27
  • $\begingroup$ @SahibaArora Hah! Yes!. Sorry. $\endgroup$ – pozcukushimatostreet Aug 15 '17 at 17:29
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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.

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    $\begingroup$ 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. $\endgroup$ – Ethan Bolker Aug 15 '17 at 17:41
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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.

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