# Proof

### Proof "absolute convergent" $\Rightarrow$ "convergent"

Let $$\sum_{k=0}^\infty x_k$$ be an absolutely convergent series. Since, be definition, $$\sum_{k=0}^\infty |x_k|$$ is convergent, we can apply Cauchy's general criterion for convergent series and conclude that for every $$\epsilon > 0$$ there is an index $$N(\epsilon)\in\mathbb N$$ such that

$\left|\sum_{k=m}^n |x_k|\right| < \epsilon\quad\quad \text{for all}\quad n\ge m\ge N(\epsilon).$

According to the triangle inequality, we have that

$\left|\sum_{k=m}^n x_k\right| \le \left|\sum_{k=m}^n |x_k|\right| < \epsilon\quad\quad \text{for all}\quad n\ge m\ge N(\epsilon).$

Applying Cauchy's general criterion for convergent series once again, we get that the series $$\sum_{k=0}^\infty x_k$$ is convergent (in the usual sense).

### Proof "convergent" $\not\Rightarrow$ "absolute convergent"

(to be done: got the proof - become a co-author?)

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### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983