3.5 Sequence

1. \(a_n = (-1)^n\), does the limit exist?

   Assume \(a_n \to L \in \mathbb{R}\). This means that \(\forall \epsilon > 0\), there exists some \(N \in \mathbb{N}\) such that \(|(-1)^n - L| < \epsilon \ \forall n \geq N\)

   Take \(\epsilon = 1\), there exists some \(N \in \mathbb{N}\) such that \(|(-1)^n - L| < 1 \ \forall n \geq N\)

   Consider \(2N > N\), then \(|(-1)^{2N} - L| < 1\) which is the same as \(|1 - L| < 1\), so \(L > 0\).

   Consider \(2N+1 > N\), then \(|(-1)^{2N+1} - L| < 1\)which is the same as \(|-1 - L| < 1\), so \(L \leq 0\).

   Contradiction!!

   So \(a_n = (-1)^n\) is not convergent.

2. Show that the sequence \(f_n(x) = \frac{x}{nx+1}\) converges uniformly to the zero function on the interval \((0,1)\)

   Let \(\epsilon > 0\), we have that \(|f_n - 0| < \epsilon \iff \frac{x}{nx+1} < \epsilon\), then \(\frac{1}{\epsilon}-\frac{1}{x}<n\)

   Since \(\frac{1}{x} > 1\), we have that \(\frac{1}{\varepsilon}-\frac{1}{x}<n\)

   So if we consider \(N > \frac{1}{\epsilon} - 1\), \(N\) does not depend on \(x\), so we get that \(\frac{x}{nx+1} < \epsilon \ \forall x \in (0,1), \forall n \geq N\)

3. Consider the sequence of functions \(f_n(x) = e^{-n^2 x^2}\) on the interval \((0, +\infty)\). Prove that the sequence converges pointwise and not uniformly.

   \(n^2 x^2 \to \infty \text{ as } n \to \infty\), so \(f_n(x) = e^{-n^2 x^2} \to 0\). Pointwise convergence to \(0\).

   Now, check if the convergence is uniform.

   \(|e^{-n^2 x^2} |<\varepsilon\iff n>\frac{\sqrt{-\log(\varepsilon)}}{x}\), but the right term goes to infinity

   Thus we cannot find such \(N\)

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