Evaluate $\lim\limits_{n \to \infty}\left(\frac{1}{1+a_1}+\frac{1}{1+a_2}+\cdots+\frac{1}{1+a_n}\right).$ Problem
Let $a_1=3,a_{n+1}=a_n^2+a_n(n=1,2,\cdots)$. Evaluate
$$\lim_{n \to \infty}\left(\frac{1}{1+a_1}+\frac{1}{1+a_2}+\cdots+\frac{1}{1+a_n}\right).$$
Attempt
Notice $$\frac{1}{1+a_n}=\frac{a_n}{a_n+a_n^2}=\frac{a_n}{a_{n+1}}$$
Then
$$\sum_{k=1}^{n}\frac{1}{1+a_k}=\sum_{k=1}^n \frac{a_k}{a_{k+1}}.$$
This will help?
 A: Prove that
$$
\sum_{k=1}^{n} \frac{1}{1 + a_{k}} = \frac{1}{3} - \frac{1}{a_{n+1}}
$$
by induction on $n$. Then the limit is $1/3$. 
A: The series is convergent.
Let's consider first the sum to $n$ terms.
$$\sum_{k=1}^{n} \frac{a_k}{a_{k+1}}$$
Multiply divide by $a_k$ and use the recursion relation for replacing $(a_k)^2 = a_{k+1} - a_{k}$. We get, 
$$\sum_{k=1}^{n} \frac{a_{k+1} - a_k}{a_{k+1}a_k} =\sum_{k=1}^{n}\left ( \frac{1}{a_{k}} -\frac{1}{a_{k+1}}\right)$$
$$= \frac{1}{a_1} - \frac{1}{a_{n+1}}$$
Now note that from the recursion relation it can be obtained that 
$$\lim_{n \to \infty} a_n \to \infty$$
Therefore the series converges to $\frac{1}{3}$.
Edit: Had made an error in the solution. Have fixed and updated now. 
A: Thanks for your enlightment. Here is a solution I've completed.
It's simple to prove an inequality $~\forall n \in \mathbb{N^+}:a_n\geq n~$. Obviously, it holds for $~n=1~$. Suppose it holds for $~n=k~$，i.e.$~a_k\geq k~$. Then
$$~a_{k+1}=a_k^2+a_k\geq k^2+k=k(k+1) \geq k+1,~$$which implies the inequality holds for $n=k+1$.Therefore, it holds for all $n$ by induction. Hence,$~a_n \to +\infty(n \to \infty)~$，and as a result, $1/a_n \to 0.$
It follows that
\begin{align*}
\lim_{n \to \infty}\sum_{k=1}^{n}\frac{1}{1+a_k}&=\lim_{n \to \infty}\sum_{k=1}^{n}\frac{a_k}{a_k+a_k^2}=\lim_{n \to \infty}\sum_{k=1}^{n}\frac{a_k}{a_{k+1}}=\lim_{n \to \infty}\sum_{k=1}^{n}\frac{a_k^2}{a_ka_{k+1}}\\
&=\lim_{n \to \infty}\sum_{k=1}^{n}\frac{a_{k+1}-a_k}{a_ka_{k+1}}=\lim_{n \to \infty}\sum_{k=1}^{n}\left(\frac{1}{a_k}-\frac{1}{a_{k-1}}\right)\\
&=\lim_{n \to \infty}\left(\frac{1}{a_1}-\frac{1}{a_{n+1}}\right)\\
&= \frac{1}{3}.
\end{align*}
