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Let $u_n$ be a sequence of positive numbers, for each n: ${u_{n+1}\over{u_n}}\le(\frac{n}{{n+1}})^\alpha$ when $\alpha>1$. Prove that $\sum_{n=1}^\infty {u_n}$ converges.

I would like to get a hint.

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2 Answers 2

up vote 2 down vote accepted

Hint: $${u_{n+1}}\le\left(\frac{n}{{n+1}} \right)^\alpha u_n$$ So we get $$u_n \le \frac{u_1}{n^\alpha},$$ and apply comparison test.

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The hypothesis implies that the sequence $n^\alpha u_n$ is non-increasing. In particular, it is bounded by $M = u_1$. Since $\alpha > 1$, this yields $$ \sum_{n=1}^\infty u_n \leq \sum_{n=1}^\infty\frac{M}{n^\alpha} < \infty $$

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