# Does the following sum converge?

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?

• 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.) – LSpice Jul 24 '20 at 18:05
$$\newcommand\an{\lfloor a n \rfloor}$$ 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 $$c_a:=a^a (1 - a)^{1 - a}\in(0,1)$$:
• 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? – Ryan Chen Jul 24 '20 at 15:18