Prove that for every positive sequence {$a_{n}$},

$$\varlimsup_{n \to \infty}\frac{\sum_{i=1}^{n+1}a_{i}}{a_{n}}\geq 4$$

Also find the sequences {$a_{n}$} for which 4 is attained.

Attempted Solution:

At the moment, I just have the following clues:

1.$$b_{n}:=\frac{\sum_{i=1}^{n+1}a_{i}}{a_{n}}, $$$$c_{n}:=\sup \left\{b_{m}\mid m\geq n\right\} , $$$$\rightarrow\varlimsup_{n \to \infty}\frac{\sum_{i=1}^{n+1}a_{i}}{a_{n}}=\lim_{n \to \infty}c_{n}$$ 2.$$b_{n}>1\rightarrow c_{n}>1\rightarrow\varlimsup_{n \to \infty}\frac{\sum_{i=1}^{n+1}a_{i}}{a_{n}}>1$$ 3. Subproblem: Is it true that for positive sequences {$a_{n}$}, $$\lim_{n \to \infty}a_{n}= \infty\to \lim_{n \to \infty} \frac{\sum_{i=1}^{n+1}a_{i}}{a_{n}}=\infty$$ If yes, then perhaps $\varlimsup_{n \to \infty}\frac{\sum_{i=1}^{n+1}a_{i}}{a_{n}}=\infty$and the result is proven for all positive unbounded sequences {$a_{n}$}.

Kindly provide me hints so that I can progress further.


Suppose that for some $N$, if $n\ge N$ $$ s_{n+1}=\sum_{k=1}^{n+1}a_k\le ca_n=c(s_n-s_{n-1})\tag{1} $$ Let $$ f(x)=\sum\limits_{k=N}^\infty s_kx^{k-N}\tag{2} $$ Then, for $x\ge0$, $$ \begin{align} \sum_{k=N}^\infty s_{k+1}x^{k-N} &\le\sum_{k=N}^\infty cs_kx^{k-N}-\sum_{k=N}^\infty cs_{k-1}x^{k-N}\\[6pt] \frac{f(x)-s_N}{x}&\le cf(x)-cxf(x)-cs_{N-1}\\[12pt] (cx^2-cx+1)f(x)&\le s_N-cs_{N-1}x\tag{3} \end{align} $$ If $c\lt4$, then $f(x)\le\frac{4s_N}{4-c}$, but since $s_k$ is an increasing sequence, $f(x)$ cannot be bounded. Thus, $c\ge4$; in other words $$ \limsup_{n\to\infty}\frac{\sum\limits_{k=1}^{n+1}a_k}{a_n}\ge4\tag{4} $$ Setting $c=4$ in inequality $(3)$ suggests we consider $$ \begin{align} f(x) &=\frac1{(1-2x)^2}\\ &=\sum_{n=0}^\infty(n+1)(2x)^n\tag{5} \end{align} $$ This leads us to notice that for sequences where $\frac{a_{n+1}}{a_n}\to2$, we get $$ \sum_{k=1}^{n+1}a_k\sim4a_n\tag{6} $$

  • $\begingroup$ nice!+1,you answer always nice! $\endgroup$ – math110 Dec 22 '13 at 15:20
  • $\begingroup$ Thanks, I'm going through it. I was about to point out a problem with the index of x in f(x) being k-N-1 but you've fixed it already! $\endgroup$ – Rhaldryn Dec 22 '13 at 15:26

Proof: Assume first that

$$\varlimsup_{n \to \infty}\dfrac{a_{n+1}}{a_{n}}=+\infty$$ then we have $$\varlimsup_{n \to \infty}\dfrac{a_{1}+a_{2}+\cdots+a_{n}+a_{n+1}}{a_{n}}=+\infty$$ Now let $$\varlimsup_{n \to \infty}\dfrac{a_{n+1}}{a_{n}}=a<+\infty$$ Then, given $\varepsilon>0$, there exists $k$ such that $$\dfrac{a_{n+1}}{a_{n}}<a+\varepsilon,n\ge k$$ In other words, $$\dfrac{a_{n}}{a_{n+1}}>\dfrac{1}{a+\varepsilon},n\ge k$$ Hence, for sufficiently large $n$, we have \begin{align*} b_{n}&=\dfrac{a_{1}+a_{2}+\cdots+a_{n}+a_{n+1}}{a_{n}}\ge\dfrac{a_{k}+\cdots+a_{n}+a_{n+1}}{a_{n}}\\ &=\dfrac{a_{k}}{a_{k+1}}\cdot\cdots\dfrac{a_{n-2}}{a_{n-1}}\cdot\dfrac{a_{n-1}}{a_{n}}+\dfrac{a_{k+1}}{a_{k+2}}\cdots\dfrac{a_{n-2}}{a_{n-1}}\cdot\dfrac{a_{n-1}}{a_{n}}\\ &+\cdots+\dfrac{a_{n-2}}{a_{n-1}}\cdot\dfrac{a_{n-1}}{a_{n}}+\dfrac{a_{n-1}}{a_{n}}+1+\dfrac{a_{n+1}}{a_{n}}\\ &\ge\left(\dfrac{1}{a+\varepsilon}\right)^{n-k}+\left(\dfrac{1}{a+\varepsilon}\right)^{n-k-1}+\cdots+\dfrac{1}{a+\varepsilon}+1+\dfrac{a_{n+1}}{a_{n}} \end{align*} if $0<a<1$, then the above inequality have $$\varlimsup_{n \to \infty}b_{n}=+\infty$$ on the other hand,if $a>1$, then we have $$\varlimsup_{n \to \infty}b_{n}=a+\lim_{n\to\infty}\dfrac{1-\left(\dfrac{1}{a+\varepsilon}\right)^{n-k+1}}{1-\dfrac{1}{a+\varepsilon}}=a+\dfrac{a+\varepsilon}{a+\varepsilon-1}$$ In case $a=1$($\varepsilon>0$ can be arbitrary) we get $$\varlimsup_{n \to \infty}b_{n}=+\infty$$ if $a>1$, then we have $$\varlimsup_{n \to \infty}b_{n}\ge 1+a+\dfrac{1}{a-1}=2+(a-1)+\dfrac{1}{a-1}\ge 4$$ $4$ is an optimal estimate because it is attained for the sequence $$a_{n}=2^n,n\in N$$

  • $\begingroup$ Nice approach (+1) $\endgroup$ – robjohn Dec 22 '13 at 14:46
  • $\begingroup$ Thanks, I'm going through it. $\endgroup$ – Rhaldryn Dec 22 '13 at 15:27

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