# 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.

Github: ### References

#### Bibliography

1. Heuser Harro: "Lehrbuch der Analysis, Teil 1", B.G. Teubner Stuttgart, 1994, 11th Edition