Does the sum $$ \lim_{n\to\infty}\sum_{k=0}^{\lfloor\alpha n \rfloor}C_n^k(-1)^k\left(1-\frac{k}{\alpha n}\right) $$ converge, where $C_n^k$ is the binomial coefficient and $0 <\alpha <1$?

The above question has been solved by Iosif Pinelis. A variation is $$ \lim_{n\to \infty}\sum_{k=0}^{\lfloor\alpha n \rfloor}C_n^k(-1)^k\left(1-\frac{k}{\alpha n}\right)^n. $$ How can we handle this sum?

  • 2
    $\begingroup$ Changing your question in response to an answer is not the preferred behaviour. You can ask a new question, but probably better first to try for yourself to see if @IosifPinelis's methods apply. (Also, the question should be self contained, and not depend on the title. I have edited accordingly, and also fixed the misspelling of @‍IosifPinelis's name.) $\endgroup$ – LSpice Jul 24 '20 at 18:05
  • 1
    $\begingroup$ Many thanks for your helpful advice. $\endgroup$ – Ryan Chen Jul 25 '20 at 0:04

$\newcommand\an{\lfloor a n \rfloor}$ Let $a:=\alpha\in(0,1)$. By induction on $m=0,1,\dots$, $$\sum_{k=0}^m \binom nk(-1)^k\Big(1-\frac k{a n}\Big) \\ =(-1)^{m+1} (a+m-a n)\frac{m+1}{an (n-1)}\,\binom n{m+1}.$$ So, letting $S_n$ denote the sum in question, we have $$S_n\sim(-1)^{\lfloor a n \rfloor+1}(a-\{a n\}) \,M_n,$$ where $\{a n\}$ is the fractional part of $a n$ and $$M_n:=\frac1n\,\binom n{\an+1}.$$ Let now $n\to\infty$. Depending on the arithmetical properties of $a$, the factor $(-1)^{\lfloor a n \rfloor+1}$ will alternate between $1$ and $-1$ and the factor $a-\{na\}$ will oscillate between $a-1<0$ and $a>0$, whereas $M_n\to\infty$, since eventually, for all large enough $n$, we have $\binom n{\an+1}\ge\min[\binom n2,\binom n{n-2}]=n(n-1)/2$. So, the sum $S_n$ will not converge to any limit.

For an illustration, here are the connected graphs $\{(n,c_a^n n^{3/2}\,S_n)\colon n=1,\dots,100\}$ for $a=1/3$ (left) and $a=\sqrt2-1$ (right), where $c_a:=a^a (1 - a)^{1 - a}\in(0,1)$:

enter image description here

  • $\begingroup$ Thanks for your answer. A variation of this sum is $\lim_{n\to \infty}\sum_{k=0}^{\lfloor\alpha n \rfloor}C_n^k(-1)^k(1-\frac{k}{\alpha n})^n$. Does this sum converge? $\endgroup$ – Ryan Chen Jul 24 '20 at 15:18
  • 1
    $\begingroup$ @RyanChen : Your question has been fully answered. If you have further questions then, for more reasons than one, I suggest you post them separately. $\endgroup$ – Iosif Pinelis Jul 24 '20 at 19:20
  • $\begingroup$ Thank you. I will first verify whether your method can be used to solve the new question. $\endgroup$ – Ryan Chen Jul 25 '20 at 0:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.