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Given the positive sequence $\{u_n\},n\in \mathbb{N}$ that meets the conditions:

$\boxed{1}$. $u_{n+1}\le u_n+u_n^2$

$\boxed{2}$. Exist the constant $\text{M} >0$ so that $\displaystyle\sum\limits_{k=1}^n u_k\le \text{M},\, \forall n\in \mathbb{N}$

Prove that $$\lim\limits_{n\to +\infty}(n\cdot u_n)=0$$

I think that we can use the Stolz-Cesaro Theorem, 0/0 Case, but I haven't found how.

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    $\begingroup$ Since there are no functions do calculate derivatives of, L'Hospital is out of the question... $\endgroup$ – 5xum Aug 31 '15 at 15:49
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    $\begingroup$ @5xum: the Cesaro-Stoltz theorem is the discrete equivalent of De L'Hopital theorem, probably the OP means that. $\endgroup$ – Jack D'Aurizio Aug 31 '15 at 15:50
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    $\begingroup$ It might help to think a little bit about why you need both conditions. Any positive decreasing sequence, say $u_n=1/n$, satisfies the first property, yet $\lim n (1/n)=1$. On the other hand, the infinite sum being finite is also not enough, because you could have a sequence which is "usually small" except that it "spikes up" occasionally, something like $u_n=1/n$ when $n=2^k$ for some $k$ and $2^{-n}$ otherwise. So somehow, if the sum is finite and $u_n$ is "eventually almost constant", you get $u_n \ll 1/n$ for large $n$. $\endgroup$ – Ian Aug 31 '15 at 16:07
  • $\begingroup$ @Ian I have a small question. Could you explain your ideas clearer in sentence "something like $u_n=1/n$ when $n=2k$ for some $k$ and $2^{−n}$ otherwise. So somehow, if the sum is finite and un is "eventually almost constant", you get $u_n \ll 1/n$ for large $n$?" Because my English is not good, so I can not understand your ideas exactly. :) $\endgroup$ – mja Sep 6 '15 at 13:08
  • $\begingroup$ @DDK Rephrasing: the first property is not enough to get your conclusion because any positive decreasing sequence will satisfy it, yet certainly there are plenty of positive decreasing sequences $u_n$ where $n u_n$ does not converge to zero. The second property is not enough to get your conclusion because I can make a convergent series where for "most" $n$, $u_n$ is much less than $1/n$, but nevertheless for infinitely many $n$, $u_n=1/n$. Then $n u_n$ will have at least two limit points, namely $0$ and $1$. $\endgroup$ – Ian Sep 6 '15 at 13:14
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Since $$ u_{n+1}\le u_n+u_n^2\tag{1} $$ we can apply the monotonically increasing function $\frac{x}{1+x}$ to both sides of $(1)$ to get $$ \frac{u_{n+1}}{1+u_{n+1}}\le\frac{u_n+u_n^2}{1+u_n+u_n^2}\le u_n\tag{2} $$ Suppose that $$ \limsup_{n\to\infty}nu_n=\varepsilon\gt0\tag{3} $$ This means that for infinitely many $n$, we have $$ u_n\ge\frac\varepsilon{2n}\tag{4} $$ For $m=\frac2\varepsilon n$, we have $u_n\ge\frac1m$, then by $(2)$, $u_{n-1}\ge\frac{\frac1m}{1+\frac1m}=\frac1{m+1}$ and by induction $$ u_n\ge\frac1m\implies u_{n-k}\ge\frac1{m+k}\tag{5} $$ thus, $$ \sum_{k=n/2}^nu_k\ge\frac{n/2}{m+n/2}=\frac\varepsilon{\varepsilon+4}\tag{6} $$ Since there are infinitely many $n$ that satisfy $(4)$, there are infinitely many intervals $\left[\frac n2,n\right]$ so that $(6)$ is true. However, then the sum of $u_n$ would diverge. Therefore, $(3)$ must be false and we must have $$ \lim_{n\to\infty}nu_n=0\tag{7} $$

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  • $\begingroup$ Elegant .I'd seen this problem before and couldn't remember how to solve it but I was sure that working back from large n would be the way. $\endgroup$ – DanielWainfleet Sep 1 '15 at 5:56
  • $\begingroup$ Very nice. Well presented too. $\endgroup$ – Colm Bhandal Sep 1 '15 at 12:44
  • $\begingroup$ The core here I feel is the use of $\frac{x}{1+x}$; what brought you to that idea? $\endgroup$ – Lord_Farin Sep 1 '15 at 16:28
  • $\begingroup$ @Lord_Farin: actually, I started by solving $u_n^2+u_n-u_{n+1}=0$ for $u_n$. The solution is $$u_n= \frac{-1+\sqrt{1+4u_{n+1}}}2= \frac{2u_{n+1}}{1+\sqrt{1+4u_{n+1}}}\ge \frac{u_{n+1}}{1+u_{n+1}}$$ Then, seeing the final result, I knew I could avoid the whole quadratic formula sidetrack by using $\frac{x}{1+x}$. $\endgroup$ – robjohn Sep 1 '15 at 17:00
  • $\begingroup$ @Lord_Farin: then, I noticed that $u_{n+1}\le u_n(1+u_n)$ has two cases: $u_{n+1}\le u_n$ and $u_{n+1}\ge u_n$ where $u_{n+1}\le u_n(1+u_{n+1})$. Either one gives $u_n\ge\frac{u_{n+1}}{1+u_{n+1}}$. $\endgroup$ – robjohn Sep 1 '15 at 17:12
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Perhaps the beginning of the solution should look like this:

Write $$n\cdot u_n=\frac{u_n}{\frac1n}$$

To apply the Cesaro-Stoltz theorem, let's try to calculate the limit $$\lim_{n\to\infty}\frac{u_{n+1}-u_n}{\frac1{n+1}-\frac1n}$$ but, applying the first condition, $$\left|\frac{u_{n+1}-u_n}{\frac1{n+1}-\frac1n}\right|\le n(n+1)u_n^2$$

But I confess that I'm stuck now, since we sholud show now that $n(n+1)u_n^2\to 0$ and I don't know how. Perhaps Cauchy Schwartz inequality combined with the condition 2?

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  • $\begingroup$ (+1) So we get that if $L=\lim n u_n$ exists, it is less than its square, so $L\leq 0$ (and that is fine since it implies $L=0$ as wanted) or $L\geq 1$, that contradicts the summability. So we only need to prove that the limit exists. $\endgroup$ – Jack D'Aurizio Aug 31 '15 at 17:35
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    $\begingroup$ If the limit can be shown to exists then we don't need Cesaro-Stoltz to conclude that the limit is $0$. A simple comparison to the harmonic series is enough. $\endgroup$ – Winther Aug 31 '15 at 17:46
  • $\begingroup$ @Winther: you are clearly right. $\endgroup$ – Jack D'Aurizio Aug 31 '15 at 17:52
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First observe that $\lim_n u_n =0$ because of hypothesis 2. Thus $$ {n \over 2}u_n $$ and $$ \left(n-\lfloor n/2\rfloor\right)u_n $$ have the same limit (if any) as $n\to\infty$. Also, the difference $\sum_{\lfloor n/2\rfloor+1}^n u_k- \left(n-\lfloor n/2\rfloor \right)u_n$ is $\sum_{\lfloor n/2\rfloor+1}^n (u_k-u_n)$, which is (by hypothesis 1.) at least $ -\sum_{\lfloor n/2\rfloor+1}^n u^2_k$. Because $\lim_n \sum_{\lfloor n/2\rfloor+1}^n u_k=0$ by hypothesis 2. and $\left(n-\lfloor n/2\rfloor \right)u_n$ is nonnegative, it suffices to show that $\lim_n\sum_{\lfloor n/2\rfloor+1}^n u^2_k =0$. But this last sum is nonnegative and at most $M\cdot\sup_{k\ge\lfloor n/2\rfloor+1}u_k$, which tends to $0$ as $n\to\infty$ because $\lim_k u_k=0$ as already noted.

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    $\begingroup$ I don't understand how you're controlling the sum of the squares. I'm also not sure that $u_k - u_n \geq -u_k^2$ always holds; what if $u_k$ is very small at some point, and then makes its way back up gradually? $\endgroup$ – Ian Aug 31 '15 at 20:45
  • $\begingroup$ Indeed the lower bound on the difference should be $-\sum_{k=\lfloor n/2\rfloor+1}^n\sum_{j=k}^{n-1} u_j^2$, and without the minus sign this latter sum of squares is at most $\sum_{j=\lfloor n/2\rfloor}^n\left(j-\lfloor n/2\rfloor\right) u_j^2$. This goes to $0$ provided $j u_j$ is bounded above by a constant... $\endgroup$ – John Dawkins Aug 31 '15 at 22:30
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Firstly, I am sorry for typo I could have, cause English is not my native language. And this is a solution of one of my friends.

Given arbitrary small number $\alpha>0$.

Supposed that the positive sequence $\{u_n\}$ has infinitely many term $u_n$ which is larger than or equal to $\alpha$.

Hence, we obtain $\displaystyle\lim_{n\to +\infty}\sum\limits_{k=1}^{n} u_k=+\infty$ which is opposite to the hypothesis $\boxed{2}$.

So, the sequence $\displaystyle\{u_n\}$ only has limited term which is larger than $\alpha$ and has infinitely many term which is less than $\alpha$. However, $\alpha$ is a arbitrary positive number, we can implies that $\displaystyle\lim_{n\to +\infty}u_n=0\tag{1}$

Hence, the sequence $\displaystyle\{n\cdot u_n\}$ either has limit $0$ or has limit $\alpha>0$.


  • Supposed that the sequence $\displaystyle\{n\cdot u_n\}$ descends to $+\infty$

Take a arbitrary number $b>\text M$. In this case, there exists a natural number $\text N_0$ so that $\displaystyle n\cdot u_n>b,\forall n>\text N_0$.

On the other hand, considering the number $\displaystyle\frac{pb}{\text N_0+p-1}$, which $p$ is a natural number. It's easily seen that $\displaystyle\frac{pb}{\text N_0+p-1}>\text M$ if $p$ is large enough.

So, if $p$ is large enough, we obtain: $$u_1+u_2+...+u_{\text N_0}+...+u_{\text N_0+p-1}>\frac{b}{\text N_0}+\frac{b}{\text N_0+1}+...+\frac{b}{\text N_0 +p-1}>\frac{pb}{\text N_0+p-1}>\text M$$ which is opposite to hypothesis of this problem. So, this case muse be eliminated.


  • Supposed that the sequence $\displaystyle\{n\cdot u_n\}$ has limit $a$: That means when $n$ descends to $+\infty$, term $u_n$ descends to $\dfrac an$. But $\displaystyle\sum_{n=1}^{\infty} \frac{a}{n}$ diverges, $\displaystyle\lim_{n\to +\infty}\sum\limits_{k=1}^{n} u_k$ descends to $+\infty$, which is opposite to hypothesis of this problem. So, this case must also be eliminated.

Conclusion, there only has a case that $\displaystyle\lim_{n\to +\infty}n.u_n=0$

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    $\begingroup$ You say: "Hence, the sequence $\displaystyle\{n\cdot u_n\}$ either has limit $0$ or has limit $\alpha>0$." Why does the sequence have to have a limit at all? Where did you use the hypothesis that $u_{n+1}\le u_n+u_n^2$? Ian's comment explains why this hypothesis is necessary. $\endgroup$ – robjohn Sep 4 '15 at 13:22
  • $\begingroup$ @robjohn Thank you. I will ask the owner of this answer. Please wait. $\endgroup$ – mja Sep 4 '15 at 14:01
  • $\begingroup$ @robjohn I have ask the owner of this solution. And his answer is: If the sequence $u_n$ doesn't meet hypothesis 1, that means there will happen the case $u_{n+1}>u_n+u_n^2=u_n(1+u_n)>u_n,\forall n$. In this case, the sequence $u_n$ must be increase, which doesn't meet hypothesis 2. $\endgroup$ – mja Sep 5 '15 at 1:18
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    $\begingroup$ That is not correct. The negation of $u_{n+1}\le u_n+u_n^2$ for all $\boldsymbol{n}$ is that $u_{n+1}\gt u_n+u_n^2$ for some $\boldsymbol{n}$. It is not the case that we are assuming $u_{n+1}\gt u_n+u_n^2$ for all $n$, so it doesn't negate hypothesis 2. $\endgroup$ – robjohn Sep 5 '15 at 3:27
  • $\begingroup$ You're right that $u_n$ must converge to zero, but that does not mean that $n u_n$ must converge at all. The second example in my first comment on the answer uses a $u_n$ where $n u_n$ has two limit points. $\endgroup$ – Ian Sep 6 '15 at 13:19

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