# Let $|\lambda|<1$. Show that $\lim_{k \rightarrow \infty} \frac{k^j}{j!}| \lambda|^{ k -j} = 0$ where $j = 1,2, \dots, m.$

Let $|\lambda|<1$. Show that $$\lim_{k \to \infty} \frac{k^j}{j!}| \lambda|^{ k -j} = 0$$ where $j = 1,2, \dots, m.$

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## 3 Answers

We have to assume that $|\lambda|<1$. Since $j$ is fixed, we have to show that $k^j|\lambda|^k$ converges to $0$ as $k\to\infty$. If you know series, you can apply ration test, otherwise, note that $\log |\lambda|<0$ (if $\lambda=0$ the problem is trivial) and $e^{-k\log|\lambda|}\geq \sum_{l=0}^{m+1}\frac{(-k\log|\lambda|)^l}{l!}$ so $|\lambda|^k\leq \left(\sum_{l=0}^{m+1}\frac{(-k\log|\lambda|)^l}{l!}\right)^{-1}$ and so $$0\leq k^j|\lambda|^k\leq k^m|\lambda|^k \leq\frac{k^m}{\sum_{l=0}^{m}\frac{(-k\log|\lambda|)^l}{l!}+\frac{(-1)^{m+1}}{(m+1)!}(k\log|\lambda|)^{m+1}}\leq \frac {(m+1)!}{k(-\log|\lambda|)^{m+1}},$$ which converges to $0$ as $k\to \infty$.

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Thank you @Davide. Nice work!! – Zizo Mar 8 '12 at 8:03

The statement wasn't true as originally stated (where we just had $\lambda < 1)$. Take for example $\lambda = -2$, $j = 1$.

If, however, you assume that $\lvert \lambda \lvert < 1$, then you can first note that (assuming $\lambda\neq 0$ since this would make the problem easy)

$\lim_{k\to\infty} \frac{k^j}{j!}\lvert \lambda\lvert^{k-j} = \frac{1}{j!}\lvert \lambda\lvert^{-j} \lim_{k\to\infty} \frac{k^j}{\lvert \lambda\lvert^{-k}}$

and then use L'Hopital's rule.

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We may assume $0<|\lambda|<1$. As $j$ is bounded by $m$ the factor $|\lambda|^{-j}$ stays bounded, so there is a constant $C$ such that $${k^j\over j!}\ |\lambda|^{k-j}\leq C\ k^m |\lambda|^k \qquad(k\geq1)\ .$$ It is well known that the right side converges to $0$ when $k\to\infty$.

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