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$$\lim_{n \to \infty} b^n n ^{\alpha} = 0, \forall |b| < 1, \alpha > 0$$

I have proved it for cases when $\alpha \le 0$, but I am not sure where to go further. I have tried D'Alembert's test, but I believe that it does not work for $\alpha > 0$.

Can you give me any direction where to go ? At least, some nominal ideas.

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Assume $\alpha > 0$, if you want to show that using the knowledge of series, consider the series $\sum_{n = 1}^\infty a_n$, where $a_n = b^n n^\alpha$.

Since $$\left|\frac{a_{n + 1}}{a_n}\right| = \left|\frac{b^{n + 1}(n + 1)^\alpha}{b^n n^\alpha}\right| = |b|\left(1 + \frac{1}{n}\right)^\alpha \to |b| \tag{$*$}$$ as $n \to \infty$. Since $|b| < 1$, the ratio test (or so-called D'Alembert test) states that this series converges. Therefore, the general term $a_n \to 0$ as $n \to \infty$.


Without using the convergence test of series, it's also straightforward to reach the conclusion. If $b = 0$, trivial. We thus assume $0 < |b| < 1$. By $(*)$ and the condition $|b| < 1$, you can see that for sufficiently large $n$, $|a_{n + 1}| < |a_n|$, so the sequence $\{|a_n|\}$ is decreasing eventually and bounded below by $0$, hence it converges to some limit $a \geq 0$, by the monotone bounded convergence theorem. Let $n \to \infty$ on both sides of $$|a_{n + 1}| = |b|\left(1 + \frac{1}{n}\right)^\alpha |a_n|,$$ we conclude that $$a = |b|a.$$ Since $0 < |b| < 1$, we must have $a = 0$.

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An approach not using any kind of convergence criterion:

We may assume that $0<b<1$ (since, $b=0$ is trivial and convergence does not depend on sign of $b$).

Then we may write

$$ b=\frac{1}{1+r} $$

where $r>0$ is real. Thus it follows that

$$ b^n=\frac{1}{(1+r)^n}=\frac{1}{1+nr+\cdots+{n \choose k} r^k+\cdots+r^n} $$

We may assume that $n>\alpha+2$ (or just for sufficiently large $n$), then choosing $k=\lfloor{\alpha}\rfloor+1=\beta$ we have

$$ b^n=\frac{1}{1+nr+\cdots+{n \choose k} r^k+\cdots+r^n}\le \frac{1}{{n \choose \beta} r^{\beta}} = \frac{1}{\frac{n(n-1)\cdots (n-\beta+1)}{\beta !} r^{\beta}} $$

so that

$$ b^n n^{\alpha} \le n^{\alpha} \frac{1}{\frac{n(n-1)\cdots (n-\beta+1)}{\beta !} r^{\beta}} \to 0 $$

as $n\to\infty$

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Because the limit of $b^n$ is $0$, since $|b| < 1$ and $n$ approaches infinity $$\lim_{n \to \infty} (b^n) (n^a) = \lim_{n \to \infty} (b^n) \cdot \lim_{n \to \infty}(n^a)$$ since limit of $b^n$ is $0$, the limit of the previous one is $0$.

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  • $\begingroup$ The logic is wrong --- you can write the limit of product as product of limits only when BOTH limits exist. $\endgroup$ – Zhanxiong Dec 7 '15 at 6:24
  • $\begingroup$ that makes sense, thanks:) sorry about that $\endgroup$ – Chuxian Wang Dec 7 '15 at 14:31

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