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

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.

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

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?