Answer to Question #156300 in Real Analysis for Tony

Question

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.

Answer 1

1. As $\forall n \in \mathbb{N}, \frac{1}{(x^2+1)^n}$ is a continuous function on $[0;1]$ and $[0;1]$ is a compact interval, $\forall n\in \mathbb{N}, I_n$ is well defined. In addition, as $\frac{1}{(1+x^2)^n} \geq 0$, the integral $I_n \geq 0$.
2. As $\frac{1}{1+x^2} \leq 1$, $\frac{1}{(1+x^2)^{n+1}} \leq \frac{1}{(1+x^2)^n}$ and thus $I_{n+1} \leq I_n$, the sequence is non-increasing. To show that it is strictly decreasing, it is enough to decompose this integral into two parts : $I_n = \int_0^{1/2}\frac{1}{(x^2+1)^n} dx + \int_{1/2}^1 \frac{1}{(x^2+1)^n} dx$, the first part is non-increasing due to the same inequality as before, in the second part $\frac{1}{x^2+1} \leq \frac{4}{5} < 1$ and thus $\frac{1}{(1+x^2)^{n+1}} < \frac{1}{(1+x^2)^n}, x\in[1/2;1]$ 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 $I_n$ is decreasing.