Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 2, Issue 3, Article 30, 2001
GOOD LOWER AND UPPER BOUNDS ON BINOMIAL COEFFICIENTS
PANTELIMON STĂNICĂ
AUBURN U NIVERSITY M ONTGOMERY,
D EPARTMENT OF M ATHEMATICS ,
M ONTGOMERY, A L 36124-4023, USA
AND
I NSTITUTE OF M ATHEMATICS OF ROMANIAN ACADEMY,
B UCHAREST-ROMANIA
stanica@strudel.aum.edu
URL: http://sciences.aum.edu/˜ stanpan
Received 6 November, 2000; accepted 26 March, 2001
Communicated by J. Sandor
A BSTRACT. We provide good bounds on binomial coefficients, generalizing known ones, using
some results of H. Robbins and of Sasvári.
Key words and phrases: Binomial Coefficients, Stirling’s Formula, Inequalities.
2000 Mathematics Subject Classification. 05A20, 11B65, 26D15.
1. M OTIVATION
Analytic techniques can be often used to obtain asymptotics for simply-indexed sequences.
Asymptotic estimates for doubly(multiply)-indexed sequences are considerably more difficult
to obtain (cf. [4], p. 204). Very little is known about how to obtain asymptotic estimates of
these sequences. The estimates that are known are based on summing over one index at a time.
For instance, according to the same source, the formula
(n−2k)2
n
2n e− 2n
p nπ
∼
k
2
3
is valid only when |2n − k| ∈ o(n 4 ).
We raise the question of getting good bounds for the binomial coefficient, which should be
valid for any n, k.
In the August-September 2000 issue of American Mathematical Monthly, O. Krafft proposed
the following problem (P10819):
ISSN (electronic): 1443-5756
c 2001 Victoria University. All rights reserved.
The author would like to thank Professor Sandor for pointing out the reference [5] and for his helpful comments.
043-00
2
PANTELIMON S TĂNIC Ă
For m ≥ 2, n ≥ 1, we have
1
mn
mm(n−1)+1
≥
n− 2 .
(m−1)(n−1)
n
(m − 1)
In this note, we are able to improve this inequality (by replacing 1 in the right-hand side by a
better absolute constant) and also generalize the inequality to mn
.
pn
We also employ a method of Sasvári [5] (see also [2]), to derive better lower and upper
bounds, with the absolute constants replaced by appropriate functions of m, n, p.
2. T HE R ESULTS
The following double inequality for the factorial
n √ was shown by H. Robbins in [3] (1955), a
step in a proof of Stirling’s formula n! ∼ ne
2πn.
Lemma 2.1 (Robbins). For n ≥ 1,
√
1
(2.1)
n! = 2π nn+ 2 e−n+r(n) ,
where r(n) satisfies
1
12n+1
< r(n) <
1
.
12n
One approach to get approximations for the binomial coefficient
use Stirling’s approximation for the factorial of Lemma 2.1, namely
√
(2.2)
1
1
2π nn+ 2 e−n+ 12n+1 < n! <
√
1
mn
pn
, m ≥ p, would be to
1
2π nn+ 2 e−n+ 12n .
Thus
mn
(mn)!
=
pn
(pn)!((m − p)n)!
>√
(2.3)
1
2π (pn)pn+ 2
√
1
1
2π (mn)mn+ 2 e−mn+ 12mn+1
√
1
1
1
e−pn+ 12pn 2π ((m − p)n)(m−p)n+ 2 e−(m−p)n+ 12n(m−p)
1
1
1
1
1
1
mmn+ 2
− 12pn
− 12n(m−p)
12nm+1
= √ n− 2
1
1 e
(m−p)n+
pn+
2p
2
2π
(m − p)
and
(2.4)
√
1
1
mn
2π (mn)mn+ 2 e−mn+ 12mn
<√
√
1
1
1
1
pn
2π (pn)pn+ 2 e−pn+ 12pn+1 2π ((m − p)n)(m−p)n+ 2 e−(m−p)n+ 12n(m−p)+1
1
1
1
1
1
1
mmn+ 2
− 12pn+1
− 12n(m−p)+1
12nm
.
= √ n− 2
1
1 e
(m−p)n+
pn+
2p
2
2π
(m − p)
However, we can improve the lower bound, by employing a method of Sasvári [5] (see also
[2]). Let
N
X
B2j
1
1
1
DN (n, m, p) =
−
−
,
2j−1
2j−1
2j−1
2j(2j
−
1)
(mn)
(np)
((m
−
p)n)
j=1
with B2j , the Bernoulli numbers defined by
∞
t
t X B2j 2j
=
1
−
+
t
et − 1
2 j=1 (2j)!
and
∆(n, m, p) = r(mn) − r(pn) − r((m − p)n).
J. Inequal. Pure and Appl. Math., 2(3) Art. 30, 2001
http://jipam.vu.edu.au/
B OUNDS ON B INOMIAL C OEFFICIENTS
3
We show that ∆(n, m, p) − DN (n, m, p) is an increasing (decreasing) function of n if N is even
(respectively, odd). We proceed to the proof of the above fact. By the Binet formula (see [2]),
we get
Z ∞ 1
t
t
r(x) =
−1+
e−tx d x, x ∈ (0, ∞),
2
t
t e −1
2
R ∞ j −t 0
and using j! = 0 t e d t, we get
Z ∞
1
∆(n, m, p) − DN (n, m, p) =
PN (t)Qn (t)d t,
t2
0
where
N
t
t X B2j 2j
PN (t) = t
−1+ −
t
e −1
2 j=1 (2j)!
and
Qn (t) = e−mnt − e(m−p)nt − e−pnt .
Sasvári proved that PN (t) is positive (negative) if N is even (respectively, odd). So we need to
show that Qn (t) is increasing with respect to n, if t > 0 and m > p ≥ 1. Since Qn (t) = f (e−nt ),
for f (u) = um − um−p − up , it suffices to show that f is decreasing on (0, 1), that is f 0 (u) < 0
on (0, 1). Now, f 0 (u) < 0 is equivalent to mum−1 − (m − p)um−p−1 − pup−1 < 0, which is
equivalent to g(u) = um−2p (mup − m + p) < p. If m ≥ 2p, then g(u) ≤ mup − m + p < p. If
1 < m < 2p, then
g 0 (u) = (m − 2p)um−2p−1 (mup − m + p) + mpum−p−1
= um−2p−1 (m − 2p)(mup − m + 2p) > 0.
Therefore, for 0 < u < 1, we have g(u) < g(1) = p and the claim is proved. Thus, we have
Theorem 2.2.
1
1
1
mmn+ 2
(2.5) √ eD2N +1 (n,m,p) n− 2
1
1
2π
(m − p)(m−p)n+ 2 ppn+ 2
1
mmn+ 2
mn
1 D2N (n,m,p) − 1
2
<
<√ e
n
1
1 .
pn
2π
(m − p)(m−p)n+ 2 ppn+ 2
Taking N = 0 and observing that B2 = 16 , we get
Corollary 2.3.
1
1
1
1
1
1
mmn+ 2
1
(2.6) √ e 12n ( m − p − m−p ) n− 2
1
1
2π
(m − p)(m−p)n+ 2 ppn+ 2
1
mmn+ 2
mn
1
− 12
<
<√ n
1
1 .
pn
2π
(m − p)(m−p)n+ 2 ppn+ 2
By using (2.4), the upper bound can be improved and we get
Corollary 2.4.
1
1
1
1
1
1
mmn+ 2
1
(2.7) √ e 12n ( m − p − m−p ) n− 2
1
1
2π
(m − p)(m−p)n+ 2 ppn+ 2
mn+ 12
1
1
1
1
1
m
mn
−
−
−
<
< √ e 12nm 12pn+1 12n(m−p)+1 n 2
1
1
pn
2π
(m − p)(m−p)n+ 2 ppn+ 2
J. Inequal. Pure and Appl. Math., 2(3) Art. 30, 2001
http://jipam.vu.edu.au/
4
PANTELIMON S TĂNIC Ă
To show that the upper bound of Corollary 2.4 improves upon the one of Corollary 2.3 we
use (2.4) and prove that
(2.8)
1
1
1
−
−
<0
12nm 12pn + 1 12n(m − p) + 1
by re-writing as
1
1
1
−
−
12nm 12pn + 1 12n(m − p) + 1
144mnp(m − p) + 12n(m − p) + 12pm + 1
=
−144mn2 (m − p) − 12mn − 144m2 np − 12mn
12mn(12pn + 1)(12n(m − p) + 1)
2
−144mnp − 12np + 12pm + 1 − 144mn2 (m − p) − 12mn
< 0.
=
12mn(12pn + 1)(12n(m − p) + 1)
Remark 2.5. The left side of Corollary 2.3 differs slightly from (2.3), in that 12mn + 1 is
replaced by 12mn. Therefore, the left side of (2.6) is an improvement of (2.3).
Next, we prove another result, where the expressions given by exponential powers are replaced by functions of n only. We prove
Theorem 2.6. Let m, n, p be positive integers, with m > p ≥ 1 and n ≥ 1. Then
1
1
1
1
mmn+ 2
(2.9) √ e− 8n n− 2
1
1
2π
(m − p)(m−p)n+ 2 ppn+ 2
<
mn
pn
1
1
1
mmn+ 2
< √ n− 2
1
1
2π
(m − p)(m−p)n+ 2 ppn+ 2
Proof. Using Corollary 2.3, we need to show that
(2.10)
1
1
1
1
−
−
≥− .
12nm 12np 12n(m − p)
8n
The inequality (2.10) is equivalent to
1
m
3
+
≤ .
m p(m − p)
2
(2.11)
Let x = m − p. Thus, x ≥ 1. We show first that the left side of (2.11), g(x, p) =
decreasing with respect to x, that is
x2 +px+p2
px(p+x)
is
d g(x, p)
1
1
=− 2 +
< 0,
dx
x
(p + x)2
which is certainly true. Therefore,
g(x, p) ≤ g(1, p) =
p2 + p + 1
(= h(p)).
p(p + 1)
Since h0 (p) = − p22p+1
< 0, we get that h is decreasing with respect to p, so
(p+1)2
3
g(x, p) ≤ h(p) ≤ h(1) = .
2
J. Inequal. Pure and Appl. Math., 2(3) Art. 30, 2001
http://jipam.vu.edu.au/
B OUNDS ON B INOMIAL C OEFFICIENTS
5
Now we provide a further simplification of Theorem 2.6. The following lemma proves to be
very useful.
m− 12
m
Lemma 2.7. Let p ≥ 1 be a fixed natural number and m ≥ p+1. Then the function m−p
is decreasing (with respect to m) and
m− 12
m
= ep .
lim
m→∞
m−p
x− 12
x
, x ≥ p + 1, is decreasing and
Proof. It suffices to prove that the function h(x) = log x−p
p
its limit is e . By differentiation
x
2xp − p
h0 (x) = log
−
.
x − p 2x(x − p)
Since
x
p
p
p2
= − log (1 − ) < + 2
log
x−p
x
x 2x
(by Taylor expansion), we get
p
p2
p
2p2 − p
x − px − p2
+ 2− −
=
< 0,
x 2x
x 2x(x − p)
x(x − p)
since p ≥ 1, so h is decreasing. The lower bound of this function is its limit, which is ep , since
x
− 12
1 − xp → e−p , and x−p
→ 1 as x → ∞.
x
h0 (x) <
Using Theorem 2.6 and Lemma 2.7, we get
Theorem 2.8. We have, for m > p ≥ 1 and n ≥ 2,
1
1
mn
1
mm(n−1)+1
> √ ep− 8n n− 2
(2.12)
1 .
pn
2π
(m − p)(m−p)(n−1)−p+1 ppn+ 2
Taking p = 1, we obtain a stronger version of the inequality P10819, namely
Corollary 2.9. We have, for m > 1 and n ≥ 2,
1
mn
mm(n−1)+1
− 8n
− 12
(2.13)
> 1.08444 e
n
.
n
(m − 1)(m−1)(n−1)
R EFERENCES
[1] O. KRAFFT, Problem P10819, Amer. Math. Monthly, 107 (2000), 652.
[2] E. RODNEY, Problem 10310, Amer. Math. Monthly, (1993), 499; with a solution in Amer. Math.
Monthly, (1996), 431–432, by MMRS.
[3] H. ROBBINS, A Remark on Stirling Formula, Amer. Math. Monthly, 62 (1955), 26–29.
[4] K. ROSEN (ed.), Handbook of Discrete Combinatorial Mathematics, CRC Press, 2000.
[5] Z. SASVÁRI, Inequalities for Binomial Coefficients, J. Math. Anal. and App., 236 (1999), 223–
226.
J. Inequal. Pure and Appl. Math., 2(3) Art. 30, 2001
http://jipam.vu.edu.au/