Determining the sum of $\frac{a_{n+1}}{a_n}$ where $a_{n+1}=\frac{na_n^2}{1+(n+1)a_n}$ 
Let $a_0=1, a_1=\frac{1}{2}, a_{n+1}=\frac{na_n^2}{1+(n+1)a_n}$, then find $\lim_{n\to\infty} \sum_{k=0}^{n}\frac{a_{k+1}}{a_k}=\dots$

We have $(a_n)$ is strictly decreasing as $a_{n+1}-a_n=\frac{-a_n(a_n+1)}{1+(n+1)a_n}<0, a_n>0$ and $ |{\frac {a_{n+1}}{a_n}|}<1$, then the sum is convergent.
Tried to telescope the sum by defining $\frac{a_{n+1}}{a_n}=\frac{na_n}{1+(n+1)a_n}=b_n$ (say). Eventually there was coming another sequence that was not giving any conclusion. Tried examining the behaviour of the terms of this sequence $$ a_0=1,a_1=\frac{1}{2},a_2=\frac{1}{2^3},a_3=\frac{1}{2^2(2^3+2+1)}\\a_4=\frac{3^2}{2^4(2^3+2^2)(2^3+3)(2^4(2^3+2^2)(2^3+2^2+1)+2^3+2^2+2+1)}$$
There would come up more $2^n$'s in the denominator for larger n's. Tried mimicking the solution as in the Convergence of $\left(a_{n+1}= \cfrac{{a_n}^2}{1+{a_n}} (n\ge 1) , a_1=1 \right)$ (seems a bit relatable), but  I am stuck here.
Any hint would be appreciated. Thanks.

You can rearrange your equation in the following manner
$a_{n+1} (1 +  (n+1) a_n) = n a_n^2$
$ na_n^2-(n+1)a_{n+1}a_n = a_{n+1}$
$\frac{a_{n+1}}{a_n} = na_n-(n+1)a_{n+1}$
The terms on the right hand side cancel out once you apply summation on both sides leaving
$\Sigma_{n=1}^{\infty} \frac{a_{n+1}}{a_n} = a_1$.
Note: Let the limit of $a_n$ as n approaches $\infty$ be $L_\infty$. You get two values for the limit: 0 and -1 by using the second equation above.
$L_\infty^2 n -(n+1)L_\infty^2= L_\infty$
$-L_\infty^2 = L_\infty$
Since all the terms are positive as you have shown $L_\infty = 0$.
Edit (Thanks to @Matija for pointing this out):
The convergence of the series $na_n$ needs to be shown instead of the series $a_n$.
