How to get the 00 result for this limit?

Consider the following series:
σ(x_):=√x−1∑k=0(x!x!)(−1)x−k(x−k)(k!(2x−k)!)+π212\sigma (\text{x$\_$})\text{:=}\sqrt{\sum _{k=0}^{x-1} \frac{(x! x!) (-1)^{x-k}}{(x-k) (k! (2 x-k)!)}+\frac{\pi ^2}{12}}
From the following codes, it seems that the series tends to 00 with a very low slope as nn approaches +∞+\infty, this is the case in the article that I’m studying

σ[x_]:=Sqrt[Pi^2/12+Sum[(x!*x!)/(k!*(2 x-k)!)*(-1)^(x-k)/(x-k),{k,0,x-1}]]

but when I compute this limit, there are the results:

(*Limit[Sqrt[π^2/12+(-1)^(2+2 n) (HarmonicNumber[n]-HarmonicNumber[2 n])],n->∞]*)

(*Sqrt[0.822467 -0.693147 2.71828^((0. +2. I) Interval[{-6.67522*10^-308,3.14159}])]*)

(*Sqrt[(0. -0.693147 I) Interval[{-1,1}]+Interval[{0.12932,1.51561}]]*)

How can I obtain 00 as the result of this limit?



2 Answers


I think you will have to ask a mathematician if the limit is really 0 (or even real) since in Mathematica you can get this

s[x_] := Sqrt[
Pi^2/12 +
Sum[(x!*x!)/(k!*(2 x – k)!)*(-1)^(x – k)/(x – k), {k, 0, x – 1}]]
s[x] // FullSimplify
Limit[s[x], x -> Infinity] // FullSimplify

So the limit seems to be complex, but I am not sure, I am not a mathematician. If you assume that you do the limit for integers, then you get this from Mathematica

Assuming[Element[x, Integers],Limit[s[x], x -> Infinity]] // FullSimplify
(*1/2 Sqrt[Ï€^2/3 – 4 Log[2]]*)

which is not 0.



Yeah, it’s pretty clear from OP’s plot that the limit isn’t zero
– JasonB
Jan 25 at 10:34



the owner of the post and @JasonB I’ve made a subtle mistake. There is k!k! in the denominator not x!x!
– sepideh
Jan 25 at 10:44



@JasonB and the original poster if I’m going to ask a mathematician,in which site do you think the question will be on-topic? in MathOverflow or in [Mathematics]?(
– sepideh
Jan 25 at 10:51



@sepideh – the question is where did the authors get that formula (eq. 6) and how does it relate to the formulas in eq. 7. before eq. 6 they say “It can be shown that the standard deviation can be evaluated in terms of n as” and then they don’t include a citation so you kind of have to take their word for it. I’ve never seen that formula for the standard deviation. But clearly the limit is not zero. I might ask over on the math SE where the authors came up with eq. 6 and how it relates to eq. 7, including this image
– JasonB
Jan 25 at 11:03



Assuming x is intended to be an integer (so we are looking at the limit of a sequence rather than a function), it reduces in the limit to 1/2 Sqrt[\[Pi]^2/3 – 4 Log[2]]. So it should be Log[2] inside the original sum, not Pi^2/12, to make the limit work out to zero.
– Daniel Lichtblau
Jan 25 at 16:59

Observing strictly that the domain of x as the upper limit of the summation index is the integers, the limit exists, it can be calculated easily with Mathematica and it is different from zero.

We need to consider this sum

\[Sigma]WH[x_] :=
Sqrt[\[Pi]^2/12 +
Simplify[Sum[(x!*x!)/(k!*(2 x – k)!)*(-1)^(x – k)/(x – k), {k, 0, x – 1}],
x \[Element] Integers]]

which is evaluated to

σWH(x)=√Hx−H2x+π212\text{$\sigma $WH}(\text{x}) = \sqrt{H_{\text{x}}-H_{2 \text{x}}+\frac{\pi ^2}{12}}

where HxH_{\text{x}} is the harmonic number of x.

The limit is then

Limit[\[Sigma]WH[x], x -> \[Infinity]]

(* Out[232]= 1/2 Sqrt[\[Pi]^2/3 – 4 Log[2]] *)

% // N

(* Out[233]= 0.359611 *)

This value is in good agreement with the asymptotic behaviour of your graph.


The essence of the agument can be studied in this simpler example

Sum[(-1)^(x – k)/(x – k), {k, 0, x – 1}]

(* Out[303]= (-1)^(2 x) ((-1)^x LerchPhi[-1, 1, 1 + x] – Log[2]) *)

Simplify[%, x \[Element] Integers]

(* Out[304]= (-1)^x LerchPhi[-1, 1, 1 + x] – Log[2] *)

Limit[%, x -> \[Infinity]]

(* Out[305]= -Log[2] *)