Show that the sequence [imath]\{x_n \}_{n\in\mathbb{N}}[/imath] defined by [math]x_n=\int_1^n \frac{\cos t}{t^2} dt[/math] is a Cauchy sequence.
I tried this: it is
[math]|x_n-x_m|=\left|\int_1^n \frac{\cos t}{t^2} dt-\int_1^m \frac{\cos t}{t^2}dt\right|=\left|\int_m^n \frac{\cos t}{t^2}dt\right| \le \int_m^n \frac{|\cos t|}{t^2} dt \le \int_m^n \frac{1}{t^2}[/math][math]=\left[-\frac{1}{t}\right]_m^n=\frac{1}{m}-\frac{1}{n}[/math]Since [imath]n\in\mathbb{N}[/imath], it is [imath]-\frac{1}{n}<0[/imath] and so [imath]\frac{1}{m}-\frac{1}{n}<\frac{1}{m}[/imath].
Now, since [imath]\frac{1}{m} \to 0[/imath] when [imath]m \to \infty[/imath], for any [imath]\epsilon>0[/imath] there exists [imath]N(\epsilon) \in \mathbb{N}[/imath] such that [imath]m \ge N(\epsilon) \implies \frac{1}{m}<\epsilon[/imath]; so, since there is no dependence from [imath]n[/imath] for [imath]\lim_{m \to \infty} \frac{1}{m}=0[/imath], it follows that for any [imath]\epsilon>0[/imath] there exists [imath]N(\epsilon) \in \mathbb{N}[/imath] such that [imath]n,m \ge N(\epsilon) \implies \frac{1}{m}<\epsilon[/imath] and so [imath]\{x_n\}_{n\in\mathbb{N}}[/imath] is a Cauchy sequence.
Is this correct?
I tried this: it is
[math]|x_n-x_m|=\left|\int_1^n \frac{\cos t}{t^2} dt-\int_1^m \frac{\cos t}{t^2}dt\right|=\left|\int_m^n \frac{\cos t}{t^2}dt\right| \le \int_m^n \frac{|\cos t|}{t^2} dt \le \int_m^n \frac{1}{t^2}[/math][math]=\left[-\frac{1}{t}\right]_m^n=\frac{1}{m}-\frac{1}{n}[/math]Since [imath]n\in\mathbb{N}[/imath], it is [imath]-\frac{1}{n}<0[/imath] and so [imath]\frac{1}{m}-\frac{1}{n}<\frac{1}{m}[/imath].
Now, since [imath]\frac{1}{m} \to 0[/imath] when [imath]m \to \infty[/imath], for any [imath]\epsilon>0[/imath] there exists [imath]N(\epsilon) \in \mathbb{N}[/imath] such that [imath]m \ge N(\epsilon) \implies \frac{1}{m}<\epsilon[/imath]; so, since there is no dependence from [imath]n[/imath] for [imath]\lim_{m \to \infty} \frac{1}{m}=0[/imath], it follows that for any [imath]\epsilon>0[/imath] there exists [imath]N(\epsilon) \in \mathbb{N}[/imath] such that [imath]n,m \ge N(\epsilon) \implies \frac{1}{m}<\epsilon[/imath] and so [imath]\{x_n\}_{n\in\mathbb{N}}[/imath] is a Cauchy sequence.
Is this correct?
