This document proves that the limit of the sum of the reciprocals of the binomial coefficients for which k is a multiple of m, as n approaches infinity, is equal to 1/m. It first considers the roots of unity to show that the sum is equal to the real part of a complex expression. It then uses De Moivre's theorem and properties of trigonometric functions to simplify the expression and show that the limit is 1/m. As an example, it explicitly proves the case where m=2.

1. Eduardo Enrique Escamilla Saldaña
Monterrey, Nuevo Leon, Mexico
February 2, 2014
Purdue Problem of the Week No.2
It is known that, for any positive integer m,
𝑛
/
𝑚𝑘
lim
!→!
!!!:!"!!
!
!!!
1
𝑛
𝑗 = 𝑚.
Prove this for m=2.
I will prove the general result. The denominator:
𝑛
𝑛
is equivalent to adding the binomial coefficients
such that
!!!:!"!!
𝑚𝑘
𝑘
𝑛
𝑘 ≡ 0 (𝑚𝑜𝑑 𝑚), namely !≡ ! (!"# !)
.
𝑘
Consider the roots of unity 𝜁 ! = cos 2𝜋𝑖𝑘 𝑚 + 𝑖 sin 2𝜋𝑖𝑘 𝑚 of 𝜁 ! − 1 = 0.
𝜁 ! − 1 = 𝜁 − 1 1 + 𝜁 + 𝜁 ! + ⋯ + 𝜁 !!!
Therefore:
!!!
𝑚 𝑖𝑓 𝑘 ≡ 0(𝑚𝑜𝑑 𝑚)
0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
𝜁 ! =
!!!
In other words:
1
𝑚
!!!
𝜁 ! =
!!!
1 𝑖𝑓 𝑘 ≡ 0(𝑚𝑜𝑑 𝑚)
0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
However by the binomial theorem:
1 + 𝑥 ! 𝜁!
1 + 𝜁!
!
!
!
=
!
=
!!!
!!!
!
!
𝑥 !! 𝜁!" Making 𝑥 = 1
𝑛 !"
𝜁
𝑘
Therefore
!≡ ! (!"# !)
𝑛
=
𝑘
𝑛
𝑘
!
!!!
1
𝑚
!!!
𝜁!
!!!
Note that
𝑒 !! + 𝑒 !!!
cos Φ =
2
2 cos Φ𝑗 = 𝑒 !!! + 𝑒 !!!!
If Φ = 2𝜋𝑗 𝑚 2 cos 2π𝑗/𝑚 = 𝑒 !
1
=
𝑚
!"!/!
!!!
1 + 𝜁!
!
!!!
+ 𝑒 !!
!"!/!
= 𝜁! + 𝜁 !!
Taking advantage of this equality
1 + 𝜁!
!
=
1
𝜁
+ 𝜁!/!
!/!
!
𝜁!/!
!
= 𝜁 !!/! + 𝜁!/!
!
𝜁!"/!
= 2 cos π𝑗/𝑚
! !"/!
𝜁

2. Since we are interested in real numbers we can take the real part of 𝜁!"/! , and using
De Moivre's theorem we get
𝑛𝑗𝜋
𝑅𝑒 𝜁!"/! = cos 2𝜋𝑗𝑛/(2𝑛) = cos
𝑚
!≡ ! (!"# !)
!!!
=
1
𝑚
𝑛
=
𝑘
2 cos π𝑗/𝑚
!!!
1+1
!
!
=
!!!
!
!!!
!
𝑛
𝑘
1
𝑚
cos
!!!
𝜁
1
=
𝑚
!
!!!
𝑛𝑗𝜋
2!
=
𝑚
𝑚
!!!
!
1 + 𝜁!
!!!
!!!
!
cos π𝑗/𝑚
cos
!!!
𝑛𝑗𝜋
𝑚
𝑛 ! !!!
1 1
= 2!
𝑗
Therefore:
𝑛
/
𝑚𝑘
lim
!→!
!!!:!"!!
= lim
!→!
2!
= lim 𝑚
!≡ ! (!"# !)
!→!
= lim
!→!
1
𝑚
= 1/𝑚
!!!
!!!
!
𝑛
𝑗
!!!
!
𝑛
/
𝑘
𝑛
𝑗 .
!!!
!
cos π𝑗/𝑚
𝑛𝑗𝜋
𝑚
cos
2!
!!!
!
cos π𝑗/𝑚
𝑛𝑗𝜋
𝑚
cos
!!!
In the case where m = 2, using the above methodology, we get
lim
!→!
= lim
!!!:!"!!
!!!
!→!
1
𝑚
1
= lim
!→! 2
𝑛
/
𝑚𝑘
cos
!!!
!
!!!
π𝑗
𝑚
𝜋𝑗
cos
2
!
𝑛
𝑗 =
!!!
!
cos
𝑛𝑗𝜋
𝑚
cos
𝑛𝑗𝜋
2
!
1
𝜋𝑗
= lim ( 1 + cos
!→! 2
2
1
= lim ( 1 + 0)
!→! 2
=1/2
!!!
)