# 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 $$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n}$$.

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