1
$\begingroup$

Show that there exists some $k \in \mathbb N$ such that $2^n>n^5$ for all $n \geq k$.

I have tried calcualating using calculator and see that the assertion made in my question is true. But how can I make it precise mathematically. Please help me in doing this.

Thank you in advance.

$\endgroup$
  • $\begingroup$ what is your Problem? $\endgroup$ – Dr. Sonnhard Graubner Nov 20 '17 at 14:52
  • $\begingroup$ Actually my problem is to answer the question in a theoritical way which I have failed to do. $\endgroup$ – user251057 Nov 20 '17 at 14:55
  • 1
    $\begingroup$ Do you know some calculus? The excersise almost become a matter of applying the definition of a limit if you use the fact that $\lim_{n\to\infty} n^5 / 2^n = 0 $. $\endgroup$ – Ove Ahlman Nov 20 '17 at 14:58
1
$\begingroup$

$$a_n=2^n$$ $$b_n=n^5$$ $$a_{n+1}=2a_n$$ $$b_{n+1}=(1+\frac{1}{n})^5\cdot b_n$$ $$\frac{a_{n+1}}{b_{n+1}}=\frac{2}{(1+\frac{1}{n})^5}\cdot \frac{a_n}{b_n}$$ Now for $n>\frac{1}{2^{1/5}-1}$ we have $\frac{a_{n+1}}{b_{n+1}}>\frac{a_{n}}{b_{n}}$

Substitute the value of $n$ you got from calculation and satisfy this condition. Then you would have proved that it is valid for all $n$ greater than the value you got

$\endgroup$
0
$\begingroup$

taking the logarithm on both sides we get $$n\ln(2)-5\ln(n)>0$$ defining $$f(n)=n\ln(2)-5\ln(n)$$ and $$f'(n)=\ln(2)-\frac{5}{n}$$ Can you proceed?

$\endgroup$
0
$\begingroup$

Claim. Let $a>1$ and $m\in\Bbb N$. Then there exists $k\in\Bbb N$ such that $a^n>n^m$ for all $n>k$.

Proof. Let $b_n=\frac{a^n}{n^m}$. Our gool is to show that $b_n>1$ for all sufficiently large $n$. Pick $q$ with $1<q<a$. Consider $n\in\Bbb N$ with $n>\frac {mq}{q-1}$. Then using the Bernoulli inequality, $$ \left(1-\frac1n\right)^m\ge 1-\frac mn>\frac 1q.$$ Hence for $n>\ell:=\lceil\frac {mq}{q-1}\rceil$, $$ \frac{b_n}{b_{n-1}}=a\cdot\frac{(n-1)^m}{n^m}=a\left(1-\frac1n\right)^m>\frac aq.$$ Now let $k>\ell$. Then for $n>k$, $$ b_n>b_\ell\cdot(a/q)^{n-\ell}=b_\ell\cdot (a/q)^{k-\ell}\cdot (a/q)^{n-k}>b_\ell\cdot (a/q)^{k-\ell}.$$ Thus all that remains to show is that there exists $k$ with $$(a/q)^k>\frac{(a/q)^\ell}{b_\ell}.$$

$\endgroup$
0
$\begingroup$

Note that $2^{24}>24^5$. Indeed,\begin{align}2^{24}>24^5&\iff2^{24}>(2^3)^5\times3^5\\&\iff2^9>3^5\\&\iff512>243.\end{align}

Now, take $n\geqslant24$ and suppose that $2^n\geqslant n^5$. Then\begin{align}2^{n+1}&=2\times2^n\\&\geqslant2\times n^5\\&\geqslant\left(\frac{n+1}n\right)^5n^5\text{ (in a moment)}\\&=(n+1)^5.\end{align}And how do I know that $2\geqslant\left(\frac{n+1}n\right)^5$? That's because $n\geqslant24$ and therefore$$\left(\frac{n+1}n\right)^5=\left(1+\frac1n\right)^5\leqslant\left(\frac{25}{24}\right)^5\leqslant2.$$

$\endgroup$
  • 1
    $\begingroup$ it is $n^5$ in the question $\endgroup$ – avz2611 Nov 20 '17 at 15:08
  • $\begingroup$ @avz2611 Thanks. I've edited my answer. $\endgroup$ – José Carlos Santos Nov 20 '17 at 15:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy