Skip navigation.

Abel criteria



a series where $ a_n\geq a_{n+1}..\geq 0 $,

$$\lim_{n\to \infty}a_n=0$$

and $ (x_n)_{n\in N} $ has partial sum $ s_n=x_0+x_1+...x_n $ equal bounded. Then

$$\sum_{n=0}^{n=+\infty}a_nx_n\: is\: convergent$$

We shall prove the convergence using Cauchy criteria
For $ \forall \epsilon>0  $ there is a $ n_{\epsilon}\in N $ such as $ \forall n\geq n_{\epsilon} $ and $ \forall p\geq 1 $we have

$$|a_{n+1}x_{n+1}+a_{n+2}x_{n+2}+.....a_{n+p}x_{n+p}|\le \epsilon$$

We have $ x_{n+i}=s_{n+i}-s_{n+i-1} $ and as $ s_n=x_0+x_1+...x_n $ are equal bounded there is a constant M such as $ \forall n\:|s_n|\leq M $
We can write

$$=|-a_{n+1}s_{n}+s_{n+1}(a_{n+1}-a_{n+2})+s_{n+2}(a_{n+2}-a_{n+3})+...s_{n++p-1}(a_{n+p-1}-a_{n+p})+s_{n+p}a_{n+p}|\leq 2Ma_{n+1}<\epsilon$$

if n is such as


Using this criteria we have the convergence of a series

$$\sum_{n=1}^{n=\infty}\frac{1}{n^{\alpha}}\cos nx $$


$$\sum_{n=1}^{n=\infty}\frac{1}{n^{\alpha}} \sin nx $$

for all $ \alpha>0 $ and $ x\neq 2k\pi $