Let $(a_n){n \ge 1}$ be a sequence of positive real numbers defined by $a_1 = \frac{1}{2}$ and $a{n + 1} = a^2n + a_n$, for every $n \in \mathbb{N}$. Prove that $\sum{k \ge 1} \frac{1}{a_k + 1}$ converges and show that
\[\sum_{k \ge 1} \frac{1}{a_k + 1} = 2.\]$Proof$:
Let $(a_1=\tfrac12)$ and $(a_{n+1}=a_n^2+a_n)$ for $(n\ge1)$. Note that $(a_n>0)$ for all $(n)$ and $(a_{n+1}=a_n(a_n+1)>a_n)$, so $(a_n)$ is increasing and unbounded $(hence (a_n\to\infty))$.
More simply observe from $a_{n+1}=a_n(a_n+1)$ that
\[\frac{1}{a_{n+1}}=\frac{1}{a_n(a_n+1)}=\frac{1}{a_n}-\frac{1}{a_n+1},\]so
\[\frac{1}{a_n+1}=\frac{1}{a_n}-\frac{1}{a_{n+1}}.\]Summing from $(k=1)$ to $N$ gives the telescoping sum
\[\sum_{k=1}^N\frac{1}{a_k+1}=\frac{1}{a_1}-\frac{1}{a_{N+1}}.\]Letting $(N\to\infty)$ and using $(a_{N+1}\to\infty)$ yields.
\[\sum_{k\ge1}\frac{1}{a_k+1}=\frac{1}{a_1}=\frac{1}{1/2}=2.\]Thus the series converges and its sum is $2$.
$\square$
(I’m not sure whether the explanation above is entirely correct. If you happen to notice any mistakes, I would really appreciate it if you could leave your suggestions in the comments when you have a moment. Thank you very much for your help and corrections.)