Find the sum

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 21, 2020

edited by littlemixfan Mar 26, 2020

edited by littlemixfan Mar 26, 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

#4

#5**0 **

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

littlemixfan
Mar 21, 2020

#9**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.

Guest Mar 23, 2020