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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 $n \geq 2$. 

 

Prove that 

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

 

for all $n \geq 1$. 

 

How do I get started with this problem? I used their hint that told me to write the relationship as $a_n = a_{n - 1} (a_{n - 1} + 1)$ and to try getting fractions but I'm not sure how to get started on that.

 

Any help would be appreciated. 

 Jun 9, 2021
 #1
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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.

 Jun 10, 2021
 #2
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Thanks Guest!

 

I appreciate the help but I already solved it by myself cheeky

xCorrosive  Jun 10, 2021

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