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Prove that $$\varlimsup_{n\rightarrow \infty} (u_n)^{\frac{1}{n}}=1,$$ where $u_0=1$ and $$u_{n+1}=\frac{2u_n^3+2u_n^2+u_n}{2u_n^2+3u_n+1}\;.$$

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2 Answers 2

up vote 4 down vote accepted

$u_n$ is clearly a non-negative sequence, and is decreasing since $$ \frac{u_{n+1} }{u_n} = \frac{2u^2_n + 2u_n +1 }{2u^2_n + 3u_n +1 } = 1 - \frac{u_n}{2u^2_n + 3u_n +1 }\leq 1.$$

Thus $u_n$ tends to some finite limit $L$ which must satisfy $\displaystyle L = \frac{2L^3+ 2L^2+ L}{2L^2 + 3L+1} .$ Thus $L=0$ and $$ \lim_{n\to\infty} \sqrt[n]{u_n} = \lim_{n\to\infty} \frac{u_{n+1}}{u_n} = 1. $$

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Thanks Ragib Zaman for your comments. I think that the solution of Alex Becker is a basic solution. –  impartialmale Mar 1 '12 at 21:13
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@impartialmale My solution is very elementary, it uses nothing beyond what students normally see in the first few weeks in a first course in analysis. If you are having trouble understanding a part of my solution, I will be happy to elaborate. –  Ragib Zaman Mar 2 '12 at 6:05

We can do even better: $\lim\limits_{n\to\infty} (u_n)^{1/n}=1$. Note that for $x>0$ we have $$\frac{d}{dx}\frac{2x^3+2x^2+x}{2x^2+3x+1}=\frac{4x^4+12x^3+10x^2+4x+1}{(2x^2+3x+1)^2}>0$$ thus since clearly each $u_n>0$, $\frac{2u_n^3+2u_n^2+u_n}{2u_n^2+3u_n+1}$ is an increasing function of $u_n$. This allows us to show that $u_n\geq \frac{1}{n+1}$ for all $n\in\mathbb N$, as $u_0=\frac{1}{1}$ and if $u_n\geq \frac{1}{n+1}$ then $$\begin{eqnarray} u_{n+1}&=&\frac{2u_n^3+2u_n^2+u_n}{2u_n^2+3u_n+1}\\ &\geq& \frac{2/(n+1)^3+2/(n+1)^2+1/(n+1)}{2/(n+1)^2+3/(n+1)+1}\\ &\geq&\frac{1}{n+1}-\frac{1}{(n+1)^2+3(n+1)+2}\\ &\geq& \frac{1}{n+1}-\frac{1}{(n+1)^2+(n+1)}\\ &\geq&\frac{1}{n+2}\end{eqnarray}$$ and so by induction $u_n\geq \frac{1}{n+1}$ for all $n$. Since it is clear that $(u_{n})^{1/n}\leq 1$ for all $n$, we need only show that $\lim\limits_{x\to\infty}\left(\frac{1}{x+1}\right)^{1/x}=1$, or equivalently (since $\ln x$ is continuous) that $$\lim\limits_{x\to\infty}\ln\left(\left(\frac{1}{x+1}\right)^{1/x}\right)=\ln 1=0$$ which we can do using L'Hospital's rule $$\lim\limits_{x\to\infty}\ln\left(\left(\frac{1}{x+1}\right)^{1/x}\right)=\lim\limits_{x\to\infty}\frac{1}{x}\ln\left(\frac{1}{x+1}\right)=\lim\limits_{x\to\infty}\frac{-1}{x+1}=0$$ and so desired result follows from squeezing $\lim\limits_{n\to\infty}(u_n)^{1/n}$ between $\lim\limits_{n\to\infty} \left(\frac{1}{n+1}\right)^{1/n}$ and $\lim\limits_{n\to\infty} 1$.

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Excellent Solution. Thanks Alex Becker very much. –  impartialmale Mar 1 '12 at 5:51
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@impartialmale Shameless self-plug: you can accept my answer by clicking on the little check-mark to the right of it (below the up/down votes). –  Alex Becker Mar 1 '12 at 6:09

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