A strictly increasing map that is not one-to-one

Consider two partially ordered sets \((E,\le)\) and \((F,\le)\) and a strictly increasing map \(f : E \to F\). If the order \((E,\le)\) is total, then \(f\) is one-to-one. Indeed for distinct elements \(x,y \in E\), we have either \(x < y\) or \(y < x\) and consequently \(f(x) < f(y)\) or \(f(y) < f(x)\). Therefore \(f(x)\) and \(f(y)\) are different. This is not true anymore for a partial order \((E,\le)\). We give a counterexample.

Consider a finite set \(E\) having at least two elements and partially ordered by the inclusion. Let \(f\) be the map defined on the powerset \(\wp(E)\) that maps \(A \subseteq E\) to its cardinal \(\vert A \vert \). \(f\) is obviously strictly increasing. However \(f\) is not one-to-one as for distincts elements \(a,b \in E\) we have \[
f(\{a\}) = 1 = f(\{b\})\]

A uniformly but not normally convergent function series

Consider a functions series \(\displaystyle \sum f_n\) of functions defined on a set \(S\) to \(\mathbb R\) or \(\mathbb C\). It is known that if \(\displaystyle \sum f_n\) is normally convergent, then \(\displaystyle \sum f_n\) is uniformly convergent.

The converse is not true and we provide two counterexamples.

Consider first the sequence of functions \((g_n)\) defined on \(\mathbb R\) by:
\[g_n(x) = \begin{cases}
\frac{\sin^2 x}{n} & \text{for } x \in (n \pi, (n+1) \pi)\\
0 & \text{else}
\end{cases}\] The series \(\displaystyle \sum \Vert g_n \Vert_\infty\) diverges as for all \(n \in \mathbb N\), \(\Vert g_n \Vert_\infty = \frac{1}{n}\) and the harmonic series \(\sum \frac{1}{n}\) diverges. However the series \(\displaystyle \sum g_n\) converges uniformly as for \(x \in \mathbb R\) the sum \(\displaystyle \sum g_n(x)\) is having only one term and \[
\vert R_n(x) \vert = \left\vert \sum_{k=n+1}^\infty g_k(x) \right\vert \le \frac{1}{n+1}\]

For our second example, we consider the sequence of functions \((f_n)\) defined on \([0,1]\) by \(f_n(x) = (-1)^n \frac{x^n}{n}\). For \(x \in [0,1]\) \(\displaystyle \sum (-1)^n \frac{x^n}{n}\) is an alternating series whose absolute value of the terms converge to \(0\) monotonically. According to Leibniz test, \(\displaystyle \sum (-1)^n \frac{x^n}{n}\) is well defined and we can apply the classical inequality \[
\displaystyle \left\vert \sum_{k=1}^\infty (-1)^k \frac{x^k}{k} – \sum_{k=1}^m (-1)^k \frac{x^k}{k} \right\vert \le \frac{x^{m+1}}{m+1} \le \frac{1}{m+1}\] for \(m \ge 1\). Which proves that \(\displaystyle \sum (-1)^n \frac{x^n}{n}\) converges uniformly on \([0,1]\).

However the convergence is not normal as \(\sup\limits_{x \in [0,1]} \frac{x^n}{n} = \frac{1}{n}\).

Root test

The root test is a test for the convergence of a series \[
\sum_{n=1}^\infty a_n \] where each term is a real or complex number. The root test was developed first by Augustin-Louis Cauchy.

We denote \[l = \limsup\limits_{n \to \infty} \sqrt[n]{\vert a_n \vert}.\] \(l\) is a non-negative real number or is possibly equal to \(\infty\). The root test states that:

  • if \(l < 1\) then the series converges absolutely;
  • if \(l > 1\) then the series diverges.

The root test is inconclusive when \(l = 1\).

A case where \(l=1\) and the series diverges

The harmonic series \(\displaystyle \sum_{n=1}^\infty \frac{1}{n}\) is divergent. However \[\sqrt[n]{\frac{1}{n}} = \frac{1}{n^{\frac{1}{n}}}=e^{- \frac{1}{n} \ln n} \] and \(\limsup\limits_{n \to \infty} \sqrt[n]{\frac{1}{n}} = 1\) as \(\lim\limits_{n \to \infty} \frac{\ln n}{n} = 0\).

A case where \(l=1\) and the series converges

Consider the series \(\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}\). We have \[\sqrt[n]{\frac{1}{n^2}} = \frac{1}{n^{\frac{2}{n}}}=e^{- \frac{2}{n} \ln n} \] Therefore \(\limsup\limits_{n \to \infty} \sqrt[n]{\frac{1}{n^2}} = 1\), while the series \(\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}\) is convergent as we have seen in the ratio test article. Continue reading Root test

Ratio test

The ratio test is a test for the convergence of a series \[
\sum_{n=1}^\infty a_n \] where each term is a real or complex number and is nonzero when \(n\) is large. The test is sometimes known as d’Alembert’s ratio test.

Suppose that \[\lim\limits_{n \to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert = l\] The ratio test states that:

  • if \(l < 1\) then the series converges absolutely;
  • if \(l > 1\) then the series diverges.

What if \(l = 1\)? One cannot conclude in that case.

Cases where \(l=1\) and the series diverges

Consider the harmonic series \(\displaystyle \sum_{n=1}^\infty \frac{1}{n}\). We have \(\lim\limits_{n \to \infty} \frac{n+1}{n} = 1\). It is well know that the harmonic series diverges. Recall that one proof uses the Cauchy’s convergence test based for \(k \ge 1\) on the inequalities: \[
\sum_{n=2^k+1}^{2^{k+1}} \frac{1}{n} \ge \sum_{n=2^k+1}^{2^{k+1}} \frac{1}{2^{k+1}} = \frac{2^{k+1}-2^k}{2^{k+1}} \ge \frac{1}{2}\]

An even simpler case is the series \(\displaystyle \sum_{n=1}^\infty 1\).

Cases where \(l=1\) and the series converges

We also have \(\lim\limits_{n \to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert = 1\) for the infinite series \(\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}\). The series is however convergent as for \(n \ge 1\) we have:\[
0 \le \frac{1}{(n+1)^2} \le \frac{1}{n(n+1)} = \frac{1}{n} – \frac{1}{n+1}\] and the series \(\displaystyle \sum_{n=1}^\infty \left(\frac{1}{n} – \frac{1}{n+1} \right)\) obviously converges.

Another example is the alternating series \(\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n}\).