User:SuperOliver

Definition
For any numerical series $$( u_n,S_n )$$, the series $$( u_n,S_n )$$ is said to 'be 'convergent' or to 'converge' if and only if the sequence (Sn) of partial sums is convergent. Otherwise the series $$( u_n,S_n )$$ is said to 'be 'divergent' or to 'diverge'.

When the series $$( u_n,S_n )$$ is convergent, we call sum of the series, denoted S, the limit of the sequence of partial sums. Thus : $$S = \lim_{n\to+\infty} S_n = \lim_{n\to+\infty} (\sum_{k=0}^n u_k)$$

We also write : $$S = \sum_{k=0}^{+\infty} u_k$$

Remark
Here again, pay attention of what you want to say ! Saying " $$( u_n )$$ converges " and " $$( u_n,S_n )$$ converges " are not two equivalent things, so, in the manipulation of these notions, in the proofs like in the exercises, avoid answers like " it converges ", which don't indicate if we are talking about the sequence $$( u_n )$$ or about the series of the general term $$ u_n $$. Be explicit in what you say in every case... and many mistakes will be avoided !

Definition
We call nature of a series the fact that the series is convergent or divergent. Two series are of the same nature if they are both convergent or both divergent.

Notation
We encounter sometimes the following notation $$\sum u_n < +\infty$$ to indicate that the series $$( u_n,S_n )$$ is convergent.

Remark
The writing $$S = \sum_{k=0}^{+\infty} u_k$$ can appear paradoxical because we do a sum of an infinite terms (second member) and we have a real finite value (first member). In reality, we mix up two close notions : what is infinite and what doesn't end. Although the two notions are close, they are not equivalent. The surface of the earth, for example, is a surface which "doesn't end" in the sense that, by walking and sailing on its surface, we never have a limit, an edge, which could prevent us from carrying on. However, the surface of the earth is not infinite. It has a finite value, which depends on its radius. The notion of series is typically in this case : we consider a sum which "doesn't end", a sum of an infinite number of terms, and we wonder if this sum is going to be finite or infinite.

How to determine the nature of a series ? A first basic answer
As explained in the paragraph above, the convergence of a series $$( u_n,S_n )$$ can be brought down to the convergence of $$( S_n )$$, the sequence of partial sums of its general term. Then, logically, in order to determine the nature of the series, it suffices : 1- to compute $$S_n$$ and to express it depending on $$n$$ 2- then to compute $$\lim_{n\to+\infty} S_n$$, which we set equal to $$S$$. a. If $$S$$ is a finite value, then $$( u_n,S_n )$$ is convergent and its sum is equal to $$S$$ b. Otherwise(i.e. $$S$$ is infinite or $$S$$ doesn't exist), $$( u_n,S_n )$$ is divergent.

It makes sense...

The unique problem is that we only know how to express $$S_n$$ depending on $$n$$ in the cases of arithmetic, geometric or telescoping series seen in the previous sections. Otherwise we don't know how to compute $$S_n$$ depending on $$n$$. And this is a true problem !

Actually, we are going to show, in the rest of this chapter, that we can determine the nature of a series $$( u_n,S_n )$$ without the need to find explicitly the expression of $$S_n$$ depending on $$n$$. To put it simply, computing the sequence of partial sums and determining the nature of the series of its general term are two aspects that we can dissociate ! We are going to see how we can do that but before that we have to go more in depth the concept of series.

First steps
Let $$( u_n,S_n )$$ be a numerical series. Let's ask ourselves the following questions to a better understand the concept of series : 1. If $$( u_n )$$ is a sequence with positive terms, how does this affect $$( S_n )$$ regarding its increasing or decreasing ? $$ S_n - S_{n-1} = u_n $$ and $$ S_n - S_{n-1} \ge 0 $$ since $$ u_n \ge 0 $$ Therefore $$ ( S_n ) $$ is an increasing sequence. 2. In this case, if $$ ( S_n ) $$ is bounded above, what can we say about the nature of the series $$( u_n,S_n )$$ ? $$ ( S_n ) $$ increases and is bounded from above $$=>$$ $$ ( S_n ) $$ is convergent $$=>$$ $$( u_n,S_n )$$ is convergent.

3. If $$( u_n )$$ is an unvarying sequence, is $$ ( S_n ) $$ an unvarying series ? No, $$ ( S_n ) $$ is increasing expect in the case where $$( u_n )$$ has a constant null value.

4. If $$ ( S_n ) $$ is an unvarying series, what does this imply on $$( u_n )$$ ? It implies that $$( u_n )$$ will be invariant in 0 over time, i.e. it will be an unvarying sequence.

5. If $$ ( S_n ) $$ strictly decreases, what does this imply on $$( u_n )$$ ? $$( u_n ) < 0$$.

6. If $$( u_n )$$ tends to a strictly positive finite limit, what is the limit of $$ ( S_n ) $$ ? $$(l>0) (\lim_{n\to+\infty} u_n = l)=> \forall n \in \N$$ $$\exists p \in \N$$ $$n \ge p =>\forall \epsilon > 0$$ $$|u_n - l| < \epsilon$$ $$=>\exists p \in \N$$ $$\forall n \ge p$$ $$\frac{l}{2} \le u_n \le \frac{3l}{2}$$ $$=>\frac{l}{2} (n-p+1) \le \sum_{k=p}^n u_k$$ $$=>\lim_{n\to+\infty} \frac{l}{2} (n-p+1) \le \lim_{n\to+\infty} \sum_{k=p}^n u_k$$ $$=>S = \lim_{n\to+\infty} S_n = +\infty$$

7. If $$( u_n )$$ tends to a strictly negative finite limit, what is the limit of $$ ( S_n ) $$ ? $$(l<0) (\lim_{n\to+\infty} u_n = l)=> \forall n \in \N$$ $$\exists p \in \N$$ $$n \ge p =>\forall \epsilon > 0$$ $$|u_n - l| < \epsilon$$ $$=>\exists p \in \N$$ $$\forall n \ge p$$ $$\frac{l}{2} \ge u_n \ge \frac{3l}{2}$$ $$=>\frac{l}{2} (n-p+1) \ge \sum_{k=p}^n u_k$$ $$=>\lim_{n\to+\infty} \frac{l}{2} (n-p+1) \ge \lim_{n\to+\infty} \sum_{k=p}^n u_k$$ $$=>S = \lim_{n\to+\infty} S_n = -\infty$$

8. Conclude on a necessary condition for the convergence of $$( S_n )$$ (hence the convergence of $$( u_n,S_n )$$). Is this condition sufficient ? It is necessary that $$\lim_{n\to+\infty} u_n = +\infty$$ for the convergence of $$( S_n )$$. This condition is not sufficient.

Theorem 5 : The necessary condition for the convergence of a series
$$( u_n,S_n )$$ converges => $$\lim_{n\to+\infty} u_n = 0$$ <=> $$\lim_{n\to+\infty} u_n \ne 0$$ => $$( u_n,S_n )$$ diverges

Proof : $$( u_n,S_n )$$ converges => $$\lim_{n\to+\infty} S_n = S$$ => $$\lim_{n\to+\infty} S_{n+1} = S$$ => $$\lim_{n\to+\infty} (S_{n+1} - S_n) = \lim_{n\to+\infty} u_{n+1} = \lim_{n\to+\infty} u_{n} = S - S = 0$$ $$\square$$

Method : To determine the nature of a series $$( u_n,S_n )$$ in particular when it has a positive general term, we always begin to examine the limit of the series $$( u_n )$$. If $$\lim_{n\to+\infty} u_n = 0$$ then the series can possibly converge. To determine for certain its nature, we need to go more depth. If $$\lim_{n\to+\infty} u_n \ne 0$$ then the series must diverge. This prerequisite, which we can quickly check, has to be done every time when we want to determine the nature of a series.

Examples : For each of these series which the general term is given hereinafter, can the series possibly converge ? 1. $$u_n = \cos({n})$$

2. $$v_k = \mathrm{Arctan}{(\frac{1}{k})}$$

3. $$w_p = \mathrm e^{-pt}$$ $$(t \in \mathbb{R}^{+*})$$ Solution : 1) $$\lim_{n\to+\infty} u_n = \lim_{n\to+\infty} \cos({n})$$ doesn't exist. It diverges.  2)$$\lim_{n\to+\infty} v_k = \lim_{n\to+\infty} \mathrm{Arctan}{(\frac{1}{k})} = 0$$. It can possibly converge. 3)$$(t \in \mathbb{R}^{+*})$$ $$w_p = \lim_{n\to+\infty} \mathrm e^{-pt} = 0$$. It can possibly converge.

Exercise 7
For each of these series which the general term is given hereinafter, compute the limit of its general term, then say if the series can possibly converge. 1. $$r_n = \frac{1-\cos({\frac{1}{n})}}{n}$$

2. $$s_n = n(1-\cos({\frac{1}{n})})$$

3. $$t_n = \frac{\sqrt{1+\frac{1}{n}}-\sqrt{1-\frac{1}{n}}}{\sqrt{2+\frac{1}{n}}-\sqrt{2-\frac{1}{n}}}$$ Solution : * $$u = \frac{1}{n} <=> n = \frac{1}{u}$$ DL(3) $$\cos(\frac{1}{n}) = \cos(u) = 1-\frac{u^2}{2!}+o(u^{3}) = 1-\frac{1}{2n^2}+o(\frac{1}{n^3})$$

1)$$\lim_{n\to+\infty} r_n = \lim_{n\to+\infty} \frac{1-1+\frac{1}{2n^2}}{n} = \lim_{n\to+\infty} \frac{1}{2n^3} = 0$$ $$( r_n,S_n )$$ can possibly converge.

2)$$\lim_{n\to+\infty} s_n = \lim_{n\to+\infty} n(1-1+\frac{1}{2n^2}) = \lim_{n\to+\infty} \frac{1}{2n} = 0$$ $$( s_n,S_n )$$ can possibly converge.

* $$(1+x)^a = 1 + ax + o(x)$$ DL(1) Therefore : $$\sqrt{1+\frac{1}{n}} = (1+\frac{1}{n})^{\frac{1}{2}} = 1 + \frac{1}{2n} + o(x)$$ $$\sqrt{1-\frac{1}{n}} = (1-\frac{1}{n})^{\frac{1}{2}} = 1 - \frac{1}{2n} + o(x)$$ $$\sqrt{2+\frac{1}{n}} = \sqrt{2(1+\frac{1}{2n}}) = \sqrt{2}\sqrt{1+\frac{1}{2n}} = \sqrt{2}(1+\frac{1}{4n} + o(x))$$ $$\sqrt{2-\frac{1}{n}} = \sqrt{2(1-\frac{1}{2n}}) = \sqrt{2}\sqrt{1-\frac{1}{2n}} = \sqrt{2}(1-\frac{1}{4n} + o(x))$$

3)$$\lim_{n\to+\infty} t_n = \lim_{n\to+\infty} \frac{(1 + \frac{1}{2n})-(1 - \frac{1}{2n})}{\sqrt{2}(1+\frac{1}{4n} - 1+\frac{1}{4n})} = \lim_{n\to+\infty} \frac{\frac{1}{n}}{\frac{\sqrt{2}}{2n}} = \lim_{n\to+\infty} \frac{2n}{\sqrt{2}n} = \frac{2\sqrt{2}}{2} = \sqrt{2} \ne 0$$ $$( t_n,S_n )$$ diverges.

Theorem 6
The set of general terms of converging numerical series forms a vector subspace of the vector space of numerical series.

Proof : Let $$( u_n )$$ and $$( v_n )$$ two sequences of positive terms such that the series $$( u_n,S_n )$$ and $$( v_n,T_n )$$ are convergent. Then the series $$( u_n+v_n,S_n+T_n )$$ is convergent and : $$\sum_{k=0}^n (u_k+v_k) =\sum_{k=0}^n u_k + \sum_{k=0}^n v_k$$ $$\forall \lambda \in \R$$, we have : $$\sum_{k=0}^n \lambda u_k = \lambda \sum_{k=0}^n u_k$$ Therefore, the set of general terms $$( u_n )$$ of converging numerical series forms a vector subspace of the vector space of numerical series. $$\square$$

Corollary 7
For all numerical series $$( u_n,S_n )$$ and $$( v_n,T_n )$$, we have :<\br> $$( u_n,S_n )$$ and $$( v_n,T_n )$$ are convergent => $$\forall \lambda \in \R$$ $$( u_n + v_n,S_n + T_n )$$ converges

Theorem 8 : Asymptotic characteristic of the convergence of a series
For all numerical series $$( u_n,S_n )$$, we have : $$(\exists p \in \N$$ $$\forall n \ge p$$ $$u_n = v_n)$$ =>($$( u_n,S_n )$$ and $$( v_n,T_n )$$ are both of the same nature) This conveys the fact that a series doesn't change its nature if we change the value of its first terms of its general term, as numerous as they may be.

Proof : $$\exists p \in \N$$ $$\forall n \ge p$$, $$0 \le \frac{1}{2} u_n \le v_n \le 2 u_n$$ $$( u_n,S_n )$$ converges => $$( 2u_n,S_n )$$ converges => $$( v_n,S_n )$$ converges $$( v_n,S_n )$$ converges => $$( \frac{1}{2}u_n,S_n )$$ converges => $$( u_n,S_n )$$ converges $$( u_n,S_n )$$ diverges => $$( \frac{1}{2}u_n,S_n )$$ diverges => $$( v_n,S_n )$$ diverges $$( v_n,S_n )$$ diverges => $$( \frac{1}{2}u_n,S_n )$$ diverges => $$( u_n,S_n )$$ diverges $$\square$$

We say that the convergence of a series is an asymptotic notion (it's related to what happening with the general term in the neighborhood of $$+\infty$$ and the values of first terms don't change anything of the convergence or the divergence). In the following sections, we are going to see rules, called "tests", which let us to determine the nature of a series. These rules enunciate at each time sufficient properties, which the general term has to verify for the convergence of $$( u_n,S_n )$$. According to this asymptotic characteristic that we have underlined, we can deduce that the set of the tests that we are going to enunciate stay applicable even if the condition(s) that $$( u_n )$$ has to verify in each of these test is/are true only from a certain rank, as great as it may be.