# Why won't Limit evaluate, and what can be done about it

I am trying to evaluate $$\lim_{n\to \infty }\sqrt{n}{{n}\choose {[np + \sqrt{np(1-p)}]}}p^{[np + \sqrt{np(1-p)}]}(1-p)^{(n-[np + \sqrt{np(1-p)}])}$$ where $0<p<1$ and $[\cdot ]$ denotes the nearest integer (if it greatly simplifies things at no cost, I don't mind if $[\cdot]$ is removed):

Limit[Sqrt[n ] Binomial[n, Round[n p + Sqrt[n p (1 - p)]]] p^
Round[n p + Sqrt[n p (1 - p)]] (1 - p)^(
n -  Round[n p - Sqrt[n p (1 - p)]] ), n -> Infinity]


Mathematica (version 11.0, Student Edition) gives the input as the answer. This limit is of the type $0\times \infty$.

In hopes of simplifying things, I removed the nearest integer function in both of the exponents, but the same issue occurs:

Limit[Sqrt[n ] Binomial[n, Round[n p + Sqrt[n p (1 - p)]]] p^(
n p + Sqrt[n p (1 - p)]) (1 - p)^(n -  n p - Sqrt[n p (1 - p)] ),
n -> Infinity, Assumptions -> 0 < p < 1]


Other Ideas: (1) I tried computing limits of products separately, but $\sqrt{n}p^{np + \sqrt{np(1-p)}}\to 0$ while the remaining product tends ComplextInfinity.

(2) I tried using bounds on the binomial coeffiecient: ${{a}\choose{b}}\le \frac{a^b}{b!}\le (\frac{a *e}{b})^b$, but the same issue occurs here when I use either of these upper bounds in place of the binomial coefficient:

Limit[Sqrt[n ] (n ^(n p + Sqrt[n p (1 - p)])/
Gamma[1 + n p + Sqrt[n p (1 - p)]]) p^(
n p + Sqrt[n p (1 - p)]) (1 - p)^(n -  n p - Sqrt[n p (1 - p)] ),
n -> Infinity, Assumptions -> 0 < p < 1]


and

Limit[Sqrt[n ] ((n E)/(n p + Sqrt[n p (1 - p)]))^(
n p + Sqrt[n p (1 - p)]) p^(n p + Sqrt[n p (1 - p)]) (1 - p)^(
n -  n p - Sqrt[n p (1 - p)] ), n -> Infinity,
Assumptions -> 0 < p < 1]


Follow up: Is there other software that is better at evaluating limits?

• Judging from the interpolation function I made of this, I don't think there's an answer to be honest. Commented Dec 31, 2016 at 13:13
• @Feyre .5? Are you getting that from your interpolation? Can you post the code for what you did? Commented Dec 31, 2016 at 13:39
• Oh, that was with later code, the original code tends clearly to 0. Just try tab = Table[{p, Sqrt[n] Binomial[n, Round[n p + Sqrt[n p (1 - p)]]] p^ Round[n p + Sqrt[n p (1 - p)]] (1 - p)^(n - Round[n p - Sqrt[n p (1 - p)]])} /. n -> 10^7, {p, 0.01, 0.99, 0.01}];,ListPlot[tab, PlotRange -> All] Commented Dec 31, 2016 at 13:48
• Make sure you seperate the two codes, first the tab=table[], then in different line the Listplot[] Commented Dec 31, 2016 at 14:02
• maybe try math.stackexchange, it might be enlightening to learn how to formally,rigorously take the limit. Commented Jan 2, 2017 at 14:29

If we remove Round, Series (around $n\to\infty$) does the trick:

$Assumptions = n \[Element] Integers && n > 2 && 0 < p < 1; Series[Sqrt[n] Binomial[n,(n p+Sqrt[n p(1-p)])] p^(n p+Sqrt[n p(1-p)])(1-p)^(n-(n p-Sqrt[n p(1-p)])), {n, Infinity, 0}] Simplify[%] // Normal Limit[%, n -> \[Infinity]] (*0*)  This works for all$0<p<1$. For$p\equiv0$, Limit[Sqrt[n] Binomial[n,(n p+Sqrt[n p(1-p)])] p^(n p+Sqrt[n p(1-p)]) (1-p)^(n-(n p-Sqrt[n p(1-p)])), p -> 0] (*Sqrt[n]*)  and therefore the limit diverges (the exact same thing happens for$p\equiv 1$). • Thank you! What do you mean by "around$n \to \infty$?" Does that mean that the center of the series is changing? If so, are we still allowed to apply the divergence test to my sequence in that case? Commented Jan 2, 2017 at 20:22 • @TheSubstitute around$n\to\infty$means that we execute Series[...,{n,∞,0}]. In more mathematical terms,$f(n)\overset{n\to\infty}=f(\infty)+\mathcal O(1/n)$. Commented Jan 2, 2017 at 20:24 expr = Sqrt[n] * Binomial[n, Round[n p + Sqrt[n p (1 - p)]]] * p^Round[n p + Sqrt[n p (1 - p)]] * (1 - p)^(n - Round[n p - Sqrt[n p (1 - p)]]); Needs["NumericalCalculus"]; Table[ {p, NLimit[expr, n -> Infinity, WorkingPrecision -> #] & /@ {15, 20, 25}} // Flatten, {p, 1/10, 9/10, 1/20}] // Grid  Increasing the WorkingPrecision produces progressively smaller values which supports the Limit being zero. • although this is very helpful, I really want to see if the limit is 0 for all$p$in$(0,1)\$. Commented Jan 2, 2017 at 8:40

Actually Limit works if you just get rid of the Round.

Limit[Sqrt[n] Binomial[n, n p + Sqrt[n p (1 - p)]]
p^(n p + Sqrt[n p (1 - p)]) (1 - p)^(n - (n p - Sqrt[n p (1 - p)])),
n -> Infinity, Assumptions -> 0 < p < 1]
`

0

Note for p near zero the limit is approached very slowly (just looking at it numerically)

• When I enter this into Mathematica (11.0, student edition), the output is the same as the input. Commented Jan 3, 2017 at 23:29
• 10.1 here, must be a version issue. Commented Jan 3, 2017 at 23:38
• It now prints that Limit cannot be followed. Commented Jul 4, 2022 at 18:32