# For $0<q_n<1, \lim\limits_{n\to\infty} q_n =q < 1$, prove $\lim\limits_{n\to\infty} n^kq_n^n = 0$ for all $k \in \Bbb N$ without L'Hôpital

I have problems with computing the following limit:

Given a sequence $0<q_n<1$ such that $\lim\limits_{n\to\infty} q_n =q < 1$, prove that for a fixed $k \in \mathbb N$, $\lim\limits_{n\to\infty} n^k q_n^n= 0$.

I know how to prove this, but I can't do it without using L'Hôpital's Rule. Does someone have an elementary proof?

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Then $q_n\leq 1-\delta$ for $n$ big enough, so that $n^k(1-\delta)^n\to0$ – Ilya Jun 6 '13 at 15:29
But How do you prove that?... – Inspiron Jun 6 '13 at 15:38
Do you know how to do this without the $n^k$ factor? It's not really clear whether you are having problems with handling $q_n$ or the $n^k$, or both (note that l'Hôpital's rule can't realistically be applied to $q_n$). – Erick Wong Jun 6 '13 at 15:50

Note that $$\frac{(n+1)^k}{n^k}=1+kn^{-1}+{k\choose2}n^{-2}+\ldots+n^{-k}\to1$$ as $n\to\infty$, hence for any $s$ with $1<s<\frac1q$ (possible because $q<1$) we can find $a$ with $n^k<a\cdot s^n$. Select $r$ with $q<r<\frac1s$ (possible because $s<\frac1q$). Then For almost all $n$, we have $q_n<r$, hence $$n^kq_n^n<n^kr^n<a(rs)^n.$$ Since $0<rs<1$, the claim follows.

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Take the positive series

$$\sum_{n=1}^\infty n^kq^n$$

apply now, say Cauchy's $\,n$-th root test:

$$\sqrt[n]{n^kq_n^n}=\left(\sqrt[n]n\right)^kq_n\xrightarrow[n\to\infty]{}1^k\cdot q=q<1$$

thus the series converges and from here the general term's sequence converges to zero.

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Since $$\lim_{n\to\infty}q_n=q<1,$$ therefore there exist such $k_0$ and $q<q'<1$ that $$q_n<q', k\geq k_0.$$ Hence $$\lim_{n\to\infty}n^kq_n^n\leq\lim_{n\to\infty}n^k(q')^n=0,$$ after root test: $$(n^k(q')^n)^{\frac{1}{n}}=(n^{\frac{1}{n}})^k q'\to q' <1.$$ Moreover $$\sum_{n=1}^{\infty}n^kq_n^n<\infty.$$

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