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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.

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  • $\begingroup$ $(4+5)^n > (4^n+5^n)$ and try similarly from the other end to show $(a)$ part. $\endgroup$ – Kirthi Raman May 12 '12 at 18:46
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    $\begingroup$ A standard way of proving that the limit is $5$ is to observe that $5^n \lt 4^n+5^n \lt (2)5^n$ and therefore $5\lt (4^n+5^n)^{1/n}\lt (2^{1/n})5$. Then since $\lim_{n\to \infty} 2^{1/n}=1$, our limit is $5$ by Squeezing. $\endgroup$ – André Nicolas May 12 '12 at 19:31
  • $\begingroup$ @Andre Thanks-) $\endgroup$ – Ryan May 12 '12 at 19:38
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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.)

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  • $\begingroup$ 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 :) $\endgroup$ – Ryan May 17 '12 at 9:41

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