Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
$(4+5)^n > (4^n+5^n)$ and try similarly from the other end to show $(a)$ part. – Kirthi Raman May 12 '12 at 18:46
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. – André Nicolas May 12 '12 at 19:31
@Andre Thanks-) – Ryan May 12 '12 at 19:38
up vote 1 down vote accepted

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


through by $5$ to obtain the equivalent inequality


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.)

share|cite|improve this answer
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 :) – Ryan May 17 '12 at 9:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.