If $u_{n+1}\le u_n+u_n^2$ and $\sum u_n$ converges, prove that $\lim\limits_{n\to +\infty}(n\cdot u_n)=0$

Question

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.

Answer 1

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 $$ \lim_{n\to\infty}nu_n=0\tag{7} $$

Answer 2

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?

Answer 3

Define $(p_n)$ and $(a_n)$ by

$$ p_n = \prod_{k=1}^{n-1} (1+u_k) \qquad\text{and}\qquad a_n = \frac{u_n}{p_n}. $$

We make the following observations:

• $\boxed{1}$ implies that $a_{n+1} \leq a_n$. That is, $(a_n)$ is positive and non-increasing.

• Since $a_n \leq u_n$, $\boxed{2}$ implies that $\sum_{n=1}^{\infty} a_n < \infty$. Then a standard argument tells that $$ \bbox[border:1px dotted navy; color:navy; padding:5px;]{ \lim_{n\to\infty} n a_n = 0. } \tag{1} $$ Indeed, monotonicity of $(a_n)$ yields $\frac{1}{2} n a_n \leq \sum_{k=\lfloor n/2 \rfloor}^{n} a_k $, which vanishes as $n \to \infty$ by $\boxed{2}$.

• By the inequality $1+x \leq e^x$, we get $p_n \leq e^{\sum_{k=1}^{n-1} u_k}$. So, $\boxed{2}$ implies $(p_n)$ is bounded above. Moreover, $(p_n)$ is non-decreasing. Hence, $$ \bbox[border:1px dotted navy; color:navy; padding:5px;]{ (p_n) \text{ converges.} } \tag{2} $$

Combining $\text{(1)}$ and $\text{(2)}$ altogether, we conclude:

$$ nu_n = na_n \cdot p_n \to 0 $$

Answer 4

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.

Answer 5

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 $\alpha>0$.

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• 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.

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• 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.

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