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I cant find an easy way to solve this, can you help?

$(2/3)^n < 1/n$

$n$ is natural and district $(1,2,3...)$

I tried using induction and basic algebra.

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closed as off-topic by user21820, pjs36, Behrouz Maleki, Adam Hughes, user223391 Dec 10 '16 at 23:50

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  • 3
    $\begingroup$ Induction is a good way to go. What stopped you? $\endgroup$ – Arthur Nov 22 '16 at 12:40
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    $\begingroup$ $(3/2)^n > n$ is a little more obvious. $\endgroup$ – Karolis Juodelė Nov 22 '16 at 12:41
  • $\begingroup$ What is $n$ is natural and district?Do you mean distinct? $\endgroup$ – tatan Nov 22 '16 at 13:16
  • $\begingroup$ @tatan it means it can only get separated values (1,2,3..) $\endgroup$ – YardenST Nov 22 '16 at 13:26
  • $\begingroup$ @YardenST That's called distinct... $\endgroup$ – tatan Nov 22 '16 at 13:31
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Induction works. The cases $n=1$ and $n=2$ are obvious. Assume that $\left(\frac{2}{3} \right)^n < \frac{1}{n}$ for $n\in \mathbf N_{>2}$. Then by the induction hypothesis $$\left(\frac{2}{3} \right)^{n+1}= \left(\frac{2}{3} \right)^n \left(\frac{2}{3} \right) < \frac{2}{3n} < \frac{1}{n+1}. $$ (The last inequality is equal to $2(n+1)<3n$ which is fulfilled because we assume $n>2$.)

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  • 1
    $\begingroup$ You only need one $<$, given by the induction hypothesis. The other $<$ can be $\le$ and you can take $n\ge 2$. $\endgroup$ – lhf Nov 22 '16 at 12:54
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By the Bernoulli inequality, $$ \left(1+\frac12\right)^n\ge 1+\frac12n. $$ As $\dfrac94>2$, ameliorate this for $n>2$ via $$ \left(\frac32\right)^n=\frac94\left(\frac32\right)^{n-2}\ge \frac94\left(1+\frac12(n-2)\right)=\frac98n>n. $$

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Hint:-

$$(\frac{2}{3})^n<\frac1n$$

$$\implies\frac{2^n}{3^n}<\frac1n$$

$$\implies n<(1.5)^n$$

and $n\in\mathbb N$.

Hence,the result follows.

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    $\begingroup$ why it follows now? n<(3/2)^n might looks right but isn't a proof $\endgroup$ – YardenST Nov 22 '16 at 21:27
  • $\begingroup$ @YardenST Now use induction....it will be easier... $\endgroup$ – tatan Nov 23 '16 at 4:24
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An alternative to LutzL's binomial answer, if $n\ge2$, then

$$\left(3\over2\right)^n=\left(1+{1\over2}\right)^n\ge1+{n\over2}+{n(n-1)\over8}={n^2+3n+8\over8}\gt n$$

The final inequality follows from

$$(n^2+3n+8)-8n=n^2-5n+8={4n^2-20n+32\over4}={(2n-5)^2+7\over4}\gt0$$

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On the LHS to go from one term to the next you multiply by 2/3. On the RHS to go from one term to the next you multiply by n/(n+1). The multiplicand on the RHS is greater than the multiplicand on the LHS for n greater than 2. Thus LHS decreases faster than RHS. It is also easily verified that LHS is less than RHS for some n. This is sufficient.

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