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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

$\frac{1}{a_1 + 1} + \frac{1}{a_2 + 1} + \dots + \frac{1}{a_n + 1} < 2$

for all $n\geq1$.

Mar 27, 2021

#1
+1

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.

Mar 27, 2021
#2
+1

Sorry, can you please put in LaTeX? Also, I don't quite understand how you got the 'quite easily' from the telescopic sum.

Guest Mar 28, 2021