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\rightarrow \infty} \frac{c_{n+1}}{c_{n}} Proof We shall prove the second inequality; the proof of the first is quite similar. Put \alpha=\limsup _{n \rightarrow \infty} \frac{c_{n+1}}{c_{n}} If a = +0 infinity, there is nothing to prove. If a is finite, choose B > a. There is an integer N such that \frac{c_{n+1}}{c_{n}} \leq \beta for n2 N. In particular, for any p>0, c_{N+k+1} \leq \beta c_{N+k} \quad(k=0,1, \ldots, p-1) Multiplying these inequalities, we obtain c_{N+p} \leq \beta^{p} c_{N} c_{n} \leq c_{N} \beta^{-N} \cdot \beta^{n} \quad(n \geq N) \sqrt[n]{c_{n}} \leq \sqrt[n]{c_{N} \beta^{-N}} \cdot \beta \underset{n \rightarrow \infty}{\lim \sup } \sqrt[n]{c_{n}} \leq \beta by Theorem 3.20(b). Since (18) is true for every B> a, we have \underset{n \rightarrow \infty}{\limsup } \sqrt[n]{c_{n}} \leq \alpha

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