ON EULER’S CONSTANT
|CALCULATING SUMS BY INTEGRALS
Li Yingying
Euler’s constant is deflned as
= limn!1Dn; (1)
where
Dn =
nX
k=1
1
k?log(n+1) (n 2N),(2)
Write
rn =?Dn,(3)
R.M,Young [1] gave an estimate of rn:
1
2(n+1) < rn <
1
2n,(4)
D.W,DeTemple [2] considered
~Dn,=
nX
k=1
1
k?log
n+ 12
·
instead of Dn and showed
7
960 ￠
1
(n+1)4 <?
~Dn + 1
24
n+ 12
·2 < 7960n4;
An earlier discussion on Dn can be found in Rippon [3],Furthermore,DeTemple and Wang [4] established
an estimate for rn with arbitrary n where Bernoulli’s numbers are involved.
In this article we give an exact expansion of rn using totally elementary method from which asymptotic
estimates for rn are derived,Our method is to calculate sums by integrals.
Rewrite Dn as
Dn =
nX
k=1
1
k?
Z k+1
k
1
xdx
!
=
nX
k=1
Z 1
0
t
k(k +t)dt,(?)
Key words and phrases,Euler constant,sums,integrals.
Mathematics Subject Classiflcation 2000,40A05
The project was supported by a grant from the Education Ministry of China
Typeset by AMS-TEX
1
2 LI YINGYING
From (?) we obtain
rn =
1X
k=n+1
Z 1
0
t
k(k +t)dt
=
1X
k=n+1
Z 1
0
t
1
k(k +t)?
1
k(k +1)
dt+
Z 1
0
tdt
1X
k=n+1
1
k(k +1)
=
1X
k=n+1
Z 1
0
t(1?t)
k(k +1)(k +t)dt+
1
n+1
Z 1
0
tdt:
Write
r1(n) =
1X
k=n+1
Z 1
0
t(1?t)
k(k +1)(k +t)dt; a1 =
Z 1
0
tdt,(5)
Then
rn = r1(n)+ a1n+1,(6)
Moreover,
r1(n) =
1X
k=n+1
Z 1
0
t(1?t)
k(k +1)(k +t)dt
=
1X
k=n+1
Z 1
0
t(1?t)
1
k(k +1)(k +t)?
1
k(k +1)(k +2)
dt
+
1X
k=n+1
Z 1
0
t(1?t)
k(k +1)(k +2)dt
=
1X
k=n+1
Z 1
0
t(1?t)(2?t)
k(k +1)(k +2)(k +t)dt
+
1X
k=n+1
1
2
Z 1
0
t(1?t)dt
1
k(k +1)?
1
(k +1)(k +2)
=
1X
k=n+1
Z 1
0
t(1?t)(2?t)
k(k +1)(k +2)(k +t)dt+
1
2
Z 1
0
t(1?t)dt 1(n+1)(n+2):
Let
r2(n) =
1X
k=n+1
Z 1
0
t(1?t)(2?t)
k(k +1)(k +2)(k +t)dt; a2 =
1
2
Z 1
0
t(1?t)dt,(7)
Then
rn = r2(n)+ a1n+1 + a2(n+1)(n+2),(8)
Deflne for m 2N; m? 2,
rm(n) =
1X
k=n+1
Z 1
0
t(1?t)(2?t)￠￠￠(m?t)
k(k +1)(k +2)￠￠￠(k +m)(k +t)dt; am =
1
m
Z 1
0
t(1?t)￠￠￠(m?1?t)dt,(9)
ON EULER’S CONSTANT |CALCULATING SUMS BY INTEGRALS 3
Then by induction we get
rn =
mX
k=1
ak
(n+1)(n+2)￠￠￠(n+k) +rm(n),(10)
From (5) and (7) we have
2a2
1X
k=n+1
1
k(k +1)2 < r1(n) < 2a2
1X
k=n+1
1
k2(k +1)
from which we derive,since a2 = 112,
1
12(n+1)(n+2) < r1(n) <
1
12n(n+1),(11)
Hence by (6) we get
1
2(n+1) +
1
12(n+1)(n+2) < rn <
1
2(n+1) +
1
12n(n+1) (12)
which is stronger than (4).
From (9) we get (for m? 2)
rm(n) < am+1
1X
k=n+1
1
(k?1)k￠￠￠(k +m?1)?
1
k(k +1)￠￠￠(k +m)
·
= am+1(n?1)!(n+m)! ; (13)
and
rm(n) >
1X
k=n+1
(m+1)am+1(k?1)!
(k +1)(k +m)! >
n(m+1)am+1
n+2
1X
k=n+1
(k?2)!
(k +m)!:
Since
1X
k=n+1
(k?2)!
(k +m)! =
1X
k=n+1
1
(k?1)k￠￠￠(k +m)
=
1X
k=n+1
1
(m+1)
1
(k?1)k￠￠￠(k +m?1)?
1
k(k +1)￠￠￠(k +m)
·
= (n?1)!(m+1)(n+m)!;
we have
rm(n) > am+1n!(n+2)(n+m)!,(14)
On the other hand,by (9) we have obviously that for m? 2
1
6m(m?2)!? am?
1
6m(m?1)!,(15)
Then we conclude for m? 2
1
6(n+2)m(m+1)?m+nm ￠ < rm(n) <
1
6n(m+1)?m+nm ￠ (16)
where m+n
m
= m!n!(m+n)!:
Taking (11) into account we see that (16) is also valid for m = 1,From (16) we get
limm!1rm(n) = 0:
Then we have established the following theorem.
4 LI YINGYING
Theorem,Let Dn = Pnk=1 1k?log(n+1) (n 2N) and = limn!1Dn be Euler constant,Then
rn,=?Dn =
1X
k=1
ak
(n+1)￠￠￠(n+k);
where
a1 = 12; ak = 1k
Z 1
0
t(1?t)￠￠￠(k?1?t)dt (k > 1):
Furthermore
1
6(n+2)m(m+1)?m+nm ￠ < rn?
mX
k=1
ak
(n+1)￠￠￠(n+k) <
1
6n(m+1)?m+nm ￠:
For the numbers ak the referee provides the following table
a1 = 12; a2 = 112; a3 = 112;a4 = 19120; a5 = 920;a6 = 863504; a7 = 1375168 ; a8 = 33953720,
He also points out that ak can be expressed in terms of Stirling numbers of the flrst kind s(k;j) as
ak = (?1)
k+1
k
kX
j=1
s(k;j)
j +1,
Our proof is completely an elementary calculation applying integral to estimate the sums,The method
may be generally applied to other cases.
Acknowledgement The author is grateful to Professor Wang who enlightened her to write down this
paper ; also,she thanks the referee for valuable suggestions and for correcting printing mistakes.
References
1,R.M.Young,Euler’s Constant,Math,Gazette 75 No.472 (1991),187{190.
2,D.W,DeTemple,A quicker convergence to Euler’s Constant,The Amer,Math,Monthly 100 (1993),468{470.
3,Rippon P L,Convergence with pictures,The Amer.Math.Monthly 93 (1986),476{478.
4,D.W,DeTemple and S.H,Wang,Half integer approximations for the partial sums of the harmonic series,J,Math.
Analysis and Applic,160 (1991),149{156.
Department of Mathematics,Beijing Normal University,Beijing 100875,P.R,China
ying-dd-li@263.net