Let $a_1=3,a_{n+1}=a_n^2+a_n(n=1,2,\cdots)$. Evaluate $$\lim_{n \to \infty}\left(\frac{1}{1+a_1}+\frac{1}{1+a_2}+\cdots+\frac{1}{1+a_n}\right).$$
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?
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.
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 $1/3$.
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*}
