How to prove the limit $\lim_{k\rightarrow\infty}\lambda^{k-j}\binom{k}{k-j}=0,\; k>j,$ and $k\gt j$

Asked Jan 20 '19 at 19:37

Active Jan 20 '19 at 20:01

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I find difficulty proving the following limit:

Suppose $|\lambda|<1$,then $$ \lim_{k\rightarrow\infty}\lambda^{k-j}\binom{k}{k-j}=0,\quad k>j. $$

When I try to prove that the $k$th Jordan matrix converges to $0$ ... (Please see this earlier question.)

edited Jan 20 '19 at 20:01

jordan_glen

asked Jan 20 '19 at 19:37

Q-Y

1

Hont. For each fixed $j$, the binomial coefficient $\binom{k}{k-j}$ is a polynomial of $k$. – Sangchul Lee Jan 20 '19 at 19:41
2

Hant. Stirling's formula may be of use, see https://en.wikipedia.org/wiki/Stirling%27s_approximation – P. Quinton Jan 20 '19 at 19:48
@SangchulLee I see it. Thank you very much. – Q-Y Jan 20 '19 at 19:49
@P.Quinton That's formula is quite useful. Thanks a lot!. – Q-Y Jan 20 '19 at 19:53
What happens when you try to prove that the kth Jordan matrix converges to $0$? – jordan_glen Jan 20 '19 at 19:53
@jordan_glen https://math.stackexchange.com/questions/326688/why-does-the-n-th-power-of-a-jordan-matrix-involve-the-binomial-coefficient I met this in the last step of the proof. – Q-Y Jan 20 '19 at 19:55
Okay, I've added a link to that question (in your question), to provide more context for this question here. Cheers! – jordan_glen Jan 20 '19 at 19:59

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