We have $\frac {1}{a_k}-\frac{1}{a_{k+1}}=\frac{1}{n+a_k}$. Sum upp these and find that $ 2-\frac{1}{a_n}=\frac{1}{n+a_0}+...+\frac{1}{n+a_{n-1}}$/ Thus, $ 2-\frac{1}{a_n}<1$ , so $ a_n<1$. So, $ a_k<1$ for all k and so $ 2-\frac{1}{a_n}>\frac{n}{n+1} $ from where it's easy to find that $ a_n>1-\frac{1}{n} $

By the property of telescoping sum, we have quite easily that $\dfrac{1}{a_n}=2-\sum_{k=0}^{n-1}\dfrac{1}{n+a_k}$. First we show that $a_n<1$ which is analogous to proving that $\sum_{k=0}^{n-1}\dfrac{1}{n+a_k}<1$. Now, we note that $\{a_k\}$ is an increasing sequence, hence $a_k\geq a_0$ for all $k\geq0$. This gives $\sum_{k=0}^{n-1}\dfrac{1}{n+a_k}\leq\dfrac{n}{n+a_0}=\dfrac{2n}{2n+1}<1$ and thus we are done. Now we prove the other part of the inequality. We note that $\sum_{k=0}^{n-1}\dfrac{1}{n+a_k}\geq\dfrac{n}{n+a_n}$ because of increasing property of the sequence $a_n$. Now using the fact that $\dfrac{1}{a_n}=2-\sum_{k=0}^{n-1}\dfrac{1}{n+a_k}$ we have, after some algebra that $2a_n^2+(n-1)a_n-n\geq0$ implying, and keeping in mind that $\{a_k\}$ is a positive sequence, $a_n\geq\dfrac{-(n-1)+\sqrt{(n-1)^2+8n}}{4}$. It remains to show that this huge quantity is greater than $1-\dfrac{1}{n}$. By squaring both sides, and cancelling out terms, we come to $3n>2$ which is true for any $n$. Hence $a_n>1-\dfrac{1}{n}$.