Proof
(related to Proposition: Raabe's Test)
- By hypothesis, \sum_{n=0}^\infty a_n is an infinite series.
- Assume, for a fixed positive number \beta > 1 there exists an index N such that a_n\neq 0 and \left|\frac{a_{n+1}}{a_n}\right|\le 1-\frac{\beta}{n} for all n\ge N.
- Now, assume there is an index N such that a_n\neq 0 and \left|\frac{a_{n+1}}{a_n}\right|\ge 1-\frac{1}{n} for all n\ge N.
- This is equivalent to n|a_{n+1}|\ge (n-1)|a_n| for all n\ge N.
- This means that the sequence (n|a_{n+1}|) is a positive-valued, monotonically increasing sequence, and therefore, its squence members exceed finally a positive bound \alpha > 0.
- It follows that \frac{|a_{n+1}|}\alpha\ge \frac 1n.
- Since the harmonic series diverges, it follows that the infinite series \frac{1}{\alpha}\sum_{n=0}^\infty a_n diverges, again by the direct comparison test.
- Therefore, except of the constant \frac{1}{\alpha}, the infinite series \sum_{n=0}^\infty a_n diverges.
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References
Bibliography
- Heuser Harro: "Lehrbuch der Analysis, Teil 1", B.G. Teubner Stuttgart, 1994, 11th Edition