+0  
 
-1
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9
avatar+93 

Find the sum 

 Mar 21, 2020
edited by littlemixfan  Mar 21, 2020
edited by littlemixfan  Mar 21, 2020
edited by littlemixfan  Mar 21, 2020
edited by littlemixfan  Mar 21, 2020
edited by littlemixfan  Mar 21, 2020
edited by littlemixfan  Mar 26, 2020
edited by littlemixfan  Mar 26, 2020
 #1
avatar+23566 
0

THAT is illegible to me...Yikes !    cheeky

 Mar 21, 2020
 #2
avatar+93 
0

Sorry, I have fixed my problem

littlemixfan  Mar 21, 2020
 #3
avatar+93 
0

Could you please help me with this question? 

littlemixfan  Mar 21, 2020
 #4
avatar+109509 
0

Even though you have edited it 4 times it is still not very legible.

 Mar 21, 2020
 #5
avatar+93 
0

I just edited it to make it as legible as possible, could your please solve this problem. thanks

littlemixfan  Mar 21, 2020
 #6
avatar+109509 
0

I have played with it but I have not gotten very far. 

Melody  Mar 21, 2020
 #7
avatar+1956 
+1

I will try but... wow! This is really complicated!!!

CalTheGreat  Mar 21, 2020
 #8
avatar+93 
+1

Thank you guys, really appreciate it!

littlemixfan  Mar 22, 2020
 #9
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0

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 23, 2020

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