2
$\begingroup$

Would someone like to help with the following question?

Prove that for $n=1,2,\ldots$

(a) $5\leq (4^n+5^n)^{1/n}\leq 10$ and that $(4^n+5^n)^{1/n}$ is bounded,

(b) $(4^n+5^n)^{1/n}\geq (4^{n+1}+5^{n+1})^{1/(n+1)}$,

(c) Hence find $\lim\limits_{n\to\infty} (4^n+5^n)^{1/n}$.

For part (a): Done.

For part (b): I have tried various methods, but am still stuck.

For part (c): (a)+(b) tells us that the given function is decreasing as n gets large, but will never become less than 5. I.e. it converges to a limit NOT LESS THAN 5. Now, I know by taking $\ln$ and then applying L'Hopital's Rule that the required limit is 5. But how to deduce that the limit is exactly 5 just from (a)+(b) alone?

Thanks in advance for any help.

  • 0
    @A$n$dre Tha$n$ks-)2012-05-12

1 Answers 1

1

André has already taken care of (c) in the comments.

In (a) you're dealing with a sequence, $\left\langle (4^n+5^n)^{1/n}:n\in\Bbb Z^+\right\rangle$, and you're to show that the inequalities hold for all positive integers $n$; it follows immediately that the sequence is bounded, since every term is between $5$ and $10$. The inequalities follow from the fact that $5^n\le 4^n+5^n\le 10^n$ for $n\in\Bbb Z^+$.

For (b) I find it convenient to divide the inequality

$(4^n+5^n)^{\frac1n}\ge(4^{n+1}+5^{n+1})^{\frac1{n+1}}$

through by $5$ to obtain the equivalent inequality

$\left(1+\left(\frac45\right)^n\right)^{\frac1n}\ge\left(1+\left(\frac45\right)^{n+1}\right)^{\frac1{n+1}}\tag{1}$

and try to prove $(1)$ instead. For notational convenience let $a=4/5$, so that we can write $(1)$ as $(1+a^n)^{1/n}\ge(1+a^{n+1})^{1/(n+1)}$.

Now let $f(x)=(1+a^x)^{1/x}$; by logarithmic differentiation we get

$\begin{align*} f\,'(x)&=(1+a^x)^{1/x}\left(\frac{\frac{xa^x\ln a}{1+a^x}-\ln(1+a^x)}{x^2}\right)\\ &=\frac1{x^2(1+a^x)^{1-1/x}}\Big(xa^x\ln a-(1+a^x)\ln(1+a^x)\Big)\;, \end{align*}$

which has the same algebraic sign as $xa^x\ln a-(1+a^x)\ln(1+a^x)$.

Clearly $1+a^x>0$, so $(1+a^x)\ln(1+a^x)>0$. But $a<1$, so $\ln a<0$, and hence $xa^x\ln a<0$ for $x>0$. Thus, $f\,'(x)<0$ for $x>0$, and $f$ is a decreasing function of $x$. This establishes $(1)$ and hence the desired inequality.

(There’s probably a nicer way, but at the moment I don’t see one.)

  • 0
    Thank you. I was able to solve (b) in 6 lines of algebraic manipulation, but the prospect of transcribing it here, what with having to figure out the laTeX for all the superscripts, iff's, and inequality symbols, seemed like such a pain that I decided to abstain. Basically I worked backwords from result (b), then expanded the result from (a) to show that my aforementioned backwords-working had resulted in a true statement, and thus (b) is proved. Perhaps future readers might like to attempt this question using this hint :)2012-05-17