+4 The sequence \(a_n \) is defined by \(a_1=\frac{1}{2}\) and \(a_n= {a}_{n-1}^{2} + {a}_{n-1}\) for \(n\geq2.\)
Prove that \(\frac{1}{a_1+1} + \frac{1}{a_2+1}+...+\frac{1}{a_n+1}<2\) for all \(n\geq1.\)
Here is what I have so far:
I rewrote the recursive sequence as: \(a_{n+1}= {a}_{n}^{2} + {a}_{n}\).
I factored the right hand side to get: \(a_{n+1}= {a}_{n}({a}_{n} + 1)\).
Then I divided both sides by \(a_n\) to get \(\frac{a_{n+1}}{a_n}= {a}_{n} + 1\).
Then I took the reciprocal of each side to get \(\frac{a_{n}}{a_{n+1}}= \frac{1}{a_n+1}\).
I am pretty sure I need to use a summation to go forward, since \(\frac{1}{a_1+1} + \frac{1}{a_2+1}+...+\frac{1}{a_n+1}\)can be rewritten as \(\sum_{k=1}^{N}\frac{1}{a_k+1}\) but am unsure how to proceed/finish this problem.
Thank you!
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, 1/(a_1 + 1) + 1/(a_2 + 1) + ... + 1/(a_n + 1) < 2.
