Question #156300
A sequence (In) is defined by Io = 1 and for every n E N* , In = Integral from 0→1 [dx / ( 1 + x^2 )^n ]
(a) Justify that In is welled defined and determine its sign.
(b) Show that In is a decreasing sequence.
1
Expert's answer
2021-02-02T05:27:53-0500
  1. As nN,1(x2+1)n\forall n \in \mathbb{N}, \frac{1}{(x^2+1)^n} is a continuous function on [0;1][0;1] and [0;1][0;1] is a compact interval, nN,In\forall n\in \mathbb{N}, I_n is well defined. In addition, as 1(1+x2)n0\frac{1}{(1+x^2)^n} \geq 0, the integral In0I_n \geq 0.
  2. As 11+x21\frac{1}{1+x^2} \leq 1, 1(1+x2)n+11(1+x2)n\frac{1}{(1+x^2)^{n+1}} \leq \frac{1}{(1+x^2)^n} and thus In+1InI_{n+1} \leq I_n, the sequence is non-increasing. To show that it is strictly decreasing, it is enough to decompose this integral into two parts : In=01/21(x2+1)ndx+1/211(x2+1)ndxI_n = \int_0^{1/2}\frac{1}{(x^2+1)^n} dx + \int_{1/2}^1 \frac{1}{(x^2+1)^n} dx, the first part is non-increasing due to the same inequality as before, in the second part 1x2+145<1\frac{1}{x^2+1} \leq \frac{4}{5} < 1 and thus 1(1+x2)n+1<1(1+x2)n,x[1/2;1]\frac{1}{(1+x^2)^{n+1}} < \frac{1}{(1+x^2)^n}, x\in[1/2;1] and so the second part is strictly decreasing. The sum of a non-increasing and a strictly secreasing sequences is a strictly decreasing sequence and thus InI_n is decreasing.

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