Example

Alternating series. A series defined by a sequence $(u_n)$ of finite real numbers is called alternating if $u_n = (-1)^n v_n$, where $v_n \geq 0$ for all $n$. Let us show that a sufficient condition for the convergence of such a series is that the sequence $(v_n)$ decreases and has 0 as its limit. For if $s_n$ denotes

the hypothesis that $(v_n)$ is decreasing implies that for all $n \geq 0$. The sequence $(s_{2n})$ [resp. $(s_{2n+1})$] is decreasing and bounded below (resp. increasing and bounded above) and therefore has a finite limit $a$ (resp. $b$), and $b \leq a$; since the assertion is proved.

If we take for example $v_n = 1/n$, the conditions above are satisfied, and therefore the series whose general term is $(-1)^n / n$ (the "alternating harmonic series") is convergent. We have seen in no. 1 that the series whose general term is $1/n$ (the "harmonic series") is not convergent, and thus the alternating harmonic series is not absolutely convergent.

We recall (Chapter III, § 5, no. 6, Proposition 7) that, if $(u_n)$ and $(v_n)$ are two convergent series of finite real numbers, then the series $(u_n + v_n)$ is convergent, and

also, if the series $(u_n)$ is convergent then the series $(au_n)$ is convergent for all finite real numbers $a$, and $\sum_{n=0}^{\infty} au_n = a \cdot \sum_{n=0}^{\infty} u_n$.

Finally, if the series $(u_n)$ and $(v_n)$ are convergent, and if $u_n \leq v_n$ for all $n$, we have $\sum_{n=0}^{\infty} u_n \leq \sum_{n=0}^{\infty} v_n$ by the principle of extension of inequalities (§ 5, no. 2, Theorem 1).

It should be noted that, if we suppose the series $(v_n)$ to be convergent but not absolutely convergent, and if $|u_n| \leq |v_n|$ for each $n$, we can by no means infer that the series $(u_n)$ is convergent, as is seen by taking $u_n = |v_n|$.

An infinite product of finite non-zero real numbers is said simply to be convergent if it is convergent in $\mathbf{R}^*$; its value is then a finite non-zero real number.