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I think I remember this from either Apostol's or Rudin's principles of Analysis.

Claim: Let $\{x_n\}_{n=1}^{\infty}$ be a sequence of positive real numbers. Then $$ \limsup \frac{x_1+x_2+...+x_n+x_{n+1}}{x_n} \geq 4.$$

Note:

  1. Yes, the $4$ is attained by $x_n=2^n$ sequence.
  2. Try $x_n=a^n$, and this holds.
  3. I have already proved that UNLESS $\lim x_n =\infty $ we will have $$ \limsup \frac{x_1+x_2+...+x_n+x_{n+1}}{x_n} = \infty.$$ So, the interesting case is when we do have a sequence going to infinity, and it is now a matter of proving that something is invariant no matter how (exponentially, linearly, polynomially, a mix of them, etc.) this convergence to infinity occurs.
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  • $\begingroup$ At the very least, we have \begin{align*}\limsup\frac{x_{1}+\cdots+x_{n+1}}{x_{n}} &=1+\limsup\frac{x_{1}+\cdots+x_{n-1}}{x_{n}}+\frac{x_{n+1}}{x_{n}} \\ &\geq2+\limsup\frac{x_{1}+\cdots+x_{n-1}}{x_{n}}\end{align*} in the divergent case. $\endgroup$
    – parsiad
    Jul 19 '16 at 1:03
  • $\begingroup$ Let me tell you something more interesting! Assume $x_{n+1}/x_n$ converges to $L$. Then $\lim \sup \frac{x_{1}+\cdots+x_{n+1}}{x_{n}} \geq \lim \frac{x_{n-1}}{x_n}+1+\lim \frac{x_{n+1}}{‌​x_{n}} \geq 1+L+\frac{1}{L} \geq 3.$ But still a $1$ is missing! $\endgroup$ Jul 19 '16 at 1:15
  • $\begingroup$ No. Even then, remember that I assumed the convergence of the fraction, which is NOT guaranteed. $\endgroup$ Jul 19 '16 at 1:24
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    $\begingroup$ This appears also as Problem 2.4.31 in Volume 1 of Kaczor, Nowak: Problems in Mathematical Analysis. It is given on page 47 with solution on page 207. $\endgroup$ Jul 19 '16 at 7:48
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Solution(as in my lecture)
Suppose that for a positive constant $c$ and a number $N$ $$ \frac{a_1+\dots a_{n+1}}{a_n} \le c \quad \text{ for all }n \ge N $$ Let $b_n:= \sum_{i=1}^n a_n$, the inequality above is equivalent to $$b_{n+1} \le c(b_{n}-b_{n-1}) \text{ for all }n \ge N$$ This form of the inequality is more or less more easier to deal with, but let's try it further, by letting $c_{n}$ denote $\frac{b_n}{b_{n-1}}$,then $$c_{n+1}+\frac{c}{c_n} \le c$$ Thus $(c_n , n \ge N)$ is a sequence of bounded real numbers, thus we can choose a subsequence $(c_{n_k},k\ge 1)$ such that $$\lim_{k \rightarrow \infty} c_{n_k}= \underbrace{\limsup_{n \rightarrow \infty } c_n}_{=:a}$$ Thus $$c \ge \limsup_{k \rightarrow \infty} \left(c_{n_k}+\frac{c}{c_{n_k-1}} \right)=a+\limsup_{k \rightarrow \infty} \frac{c}{c_{n_k-1}} \ge a+ \frac{c}{ \limsup c_n} =a+\frac{c}{a} \ge 2\sqrt{c} $$ Or, $c$ must be bigger than $4$. This means$$\text{ if} c \ge \limsup \frac{ a_1+\dots+a_{n+1}}{a_n} \text{ then } c \ge 4$$ Hence the conclusion. $\square$

A small story: This question brings back lots of memories. The first time I encounter it was when solving the famous analysis problem books by Kaczor and Nowak. A very long time ago. Then after that, in a small tutor class with gifted highschool students, I told them that it was important to find a good presentation of your problem.
And this question was the example I used in that class.

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    $\begingroup$ Very nice! There is a similar approach on AoPS (artofproblemsolving.com/community/c7h1247405p6417340), but your's is simpler. $\endgroup$
    – Martin R
    Oct 28 at 7:53
  • $\begingroup$ Thank you, Martin $\endgroup$ Oct 28 at 7:56
  • $\begingroup$ Your argument about the sequence $(c_n)$ can be generalized in the following way: Let $(x_n)$ be a sequence of positive real numbers, $A, B > 0$, and $P$ a permutation of the positive integers. If $x_n + \frac{A}{x_{P_n}} \le B$ for all $n$ then $B^2 \ge 4A$. I wonder if that is of any use ... $\endgroup$
    – Martin R
    Oct 28 at 8:14
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    $\begingroup$ You mean $ \limsup \frac{1}{c_{n_k-1}} \ge \frac{1}{ \limsup c_n}$ ? $\endgroup$ Oct 28 at 17:14
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    $\begingroup$ Thanks a lot :). I would like to add to my previous comment that the equality is valid only for positive $c_{n_{k-1}}$'s. +1 $\endgroup$
    – Koro
    Oct 28 at 17:18
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First, assume that $\lim\sup \frac{x_{1}+\cdots+x_{n+1}}{x_{n}}=L>0$ is finite. Define a sequence of polynomial $\{p_{n}(x)\}$ by $p_{0}(x)=1, p_{1}(x)=x-1$ and \begin{align} p_{n+1}(x)=(x-1)p_{n}(x)-p_{n-1}(x)-\cdots-p_{1}(x)-p_{0}(x).\,\,(n\geq 1) \end{align}

Claim. $p_{n}(L)>0$ for all $n\geq 1$.

proof. We have $x_{1}+x_{2}+\cdots+x_{n+1}\leq Lx_{n}$ for all $n\geq 1$. (Edited : Actually, we only have this for sufficiently large $n$. However, this can be modified to all $n$ with a user254665's comment.) By induction, we will prove the following : \begin{align} x_{n+1}\leq \frac{p_{n}(L)}{p_{n-1}(L)}x_{n} \end{align} (If we prove this inequality, then claim immediately follows from $x_{n+1}, x_{n}>0$.) Since $x_{1}+x_{2}\leq Lx_{1}$, it holds for $n=1$. If it holds for $n\leq N-1$, then \begin{align} x_{N+1}&\leq (L-1)x_{N}-x_{N-1}-x_{N-2}-\cdots-x_{1} \\ &\leq \left(L-1-\frac{p_{N-2}(L)}{p_{N-1}(L)}-\frac{p_{N-2}(L)}{p_{N-1}(L)}\frac{p_{N-3}(L)}{p_{N-2}(L)}-\cdots-\frac{p_{N-2}(L)}{p_{N-1}(L)}\frac{p_{N-3}(L)}{p_{N-2}(L)}\cdots \frac{p_{0}(L)}{p_{1}(L)}\right)x_{N} \\ &=\frac{p_{N}(L)}{p_{N-1}(L)}x_{N}. \end{align}

Now using the recurrence relation, we can prove $p_{n+1}(x)=x(p_{n}(x)-p_{n-1}(x))$ by subtracting two recurrence relations for $n$ and $n-1$. Then by using characteristic polynomial method, we get \begin{align} p_{n}(x)=\frac{\sqrt{x^{2}-4x}+x-2}{2\sqrt{x^{2}-4x}}\left(\frac{x+\sqrt{x^{2}-4x}}{2}\right)^{n}+\frac{\sqrt{x^{2}-4x}-x+2}{2\sqrt{x^{2}-4x}}\left(\frac{x-\sqrt{x^{2}-4x}}{2}\right)^{n} \end{align} for $x\neq 4$.

Now, suppose $0<L<4$. Let \begin{align} \frac{L+\sqrt{L^{2}-4L}}{2}=\frac{L+i\sqrt{4L-L^{2}}}{2}=\sqrt{L}e^{i\theta} \end{align} where $\cos \theta=\frac{\sqrt{L}}{2}<1$. By direct computation, we have \begin{align} p_{n}(L)=L^{n/2}\left(\cos n\theta+\frac{L-2}{\sqrt{4L-L^{2}}}\sin n\theta\right)=\frac{2L^{n/2}}{\sqrt{4L-L^{2}}} \sin (n\theta+\alpha) \end{align} where $\cos\alpha=\frac{L-2}{2}$. Since $0<\theta<\pi/2$, we can choose $n$ s.t. $\sin (n\theta+\alpha)<0\Rightarrow p_{n}(L)<0$. Contradiction. So we have $L\geq 4$.


There is an easier proof which is discovered by my friend. Let $S_{n}=x_{1}+\cdots + x_{n}$ and $t_{n}=S_{n}/S_{n+1}$. Then $0<t_{n}<1$. Suppose $\limsup\frac{x_{1}+\cdots+x_{n+1}}{x_{n}}=L<4$. Then we have \begin{align} \frac{x_{1}+\cdots+x_{n+1}}{x_{n}}\leq L\Leftrightarrow t_{n}\geq \frac{1}{L(1-t_{n-1})} \end{align} for $n\geq N$ for some $N\in \mathbb{N}$. If we put $f(x)=\frac{1}{L(1-x)}$, $t_{n}\geq f(t_{n-1})$. We can check that $f(x)>x$ and $t_{n}$ diverges, i.e. $t_{n}>1$ for some $n$. Contradiction.

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  • $\begingroup$ We can show that $p_{n}(x)=\sum_{r\geq 0} (-1)^{r}\binom{n+1-r}{r}x^{n-r}$, but I think this is not helpful to prove.. $\endgroup$
    – Seewoo Lee
    Jul 19 '16 at 5:52
  • $\begingroup$ Logical flaw : $\lim \sup \sum_1^{n+1}x_j/x_n =L$ with $0<L<\infty$ does NOT imply that $\sum_1^{n+1}x_j\leq Lx_n$ for all $n. $ This can be fixed as follows: Suppose that $0<L<4.$ Then fix $m$ such that $n>m\implies \sum_1^{n+1}x_i/x_n<4. $ Now for $ j=1,..,m $ replace $ x_j $ by $ s2^j $ where $ s $ is small enough that $ s2^j\leq x_j $ for $ j=1,...,m. $ Then the rest of your A works verbatim, arriving at the desired contradiction. $\endgroup$ Jul 20 '16 at 5:35
  • $\begingroup$ @user254665 Thank you very much. I'll edit it. $\endgroup$
    – Seewoo Lee
    Jul 22 '16 at 0:10
  • $\begingroup$ Now I'm not sure whether my fix works. There may be a problem with the value of $\sum_1^m x_i/x_m . $ $\endgroup$ Jul 22 '16 at 1:01
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    $\begingroup$ Nitpicking: $\limsup\frac{x_{1}+\cdots+x_{n+1}}{x_{n}}=L$ does not imply $\frac{x_{1}+\cdots+x_{n+1}}{x_{n}}\leq L$ for sufficiently large $N$. This can be salvaged of course by choosing $L$ greater that the limes superior, and less than $4$. $\endgroup$
    – Martin R
    Oct 28 at 7:57

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