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.

Consider the sequence $b_n=na_n$ and notice that $b_{n}-b_{n+1}=\frac{b_n}{1+\frac{n+1}{n}b_n}\in(0,b_n)$ for $n>0$, so $b_n$ is positive and decreasing, and thereby convergent. Let $b=\lim_{n\rightarrow\infty}b_n$. Combining the convergence with the above we have $0=\lim_{n\rightarrow\infty}(b_{n+1}-b_n)=\frac{b}{1+b}$, which gives $b=0$.
On the other hand, notice that $\frac{a_{n+1}}{a_n}=\frac{b_n^2}{na_n(1+(n+1)a_n)}=b_{n}-b_{n+1}$ and hence $\sum_{k=0}^n\frac{a_{n+1}}{a_n}=\frac{1}{2}+\sum_{k=1}^n(b_{k}-b_{k+1})=b_1+b_1-b_{n+1}=1-b_{n+1}$. Since we have already established that $b=0$, we have $\sum_{k=0}^\infty\frac{a_{n+1}}{a_n}=1$.
