Half-integer estimates for harmonic sums and the digamma function:
de Temple’s method
Notes by G.J.O. Jameson
Write Hn for the harmonic sum
Pn
1
r=1 r .
The following well-known estimation can be
established by Euler-Maclaurin summation, or by the logarithmic and binomial series:
1
1
Hn − γ = log n +
−
+ r(n),
(1)
2n 12n2
where γ is Euler’s constant and
1
0 ≤ r(n) ≤
.
120n4
1
1
Since log(n + 21 ) = log n + 2n
+ O( n12 ), it is a natural idea to absorb the term 2n
into the
logarithmic term and compare Hn − γ directly with log(n + 21 ). This was done by De Temple
[DT]. His result (reproduced in [BM, p. 181–2]) states that
Hn − γ = log(n + 21 ) + r1 (n),
(2)
where
1
1
≤ r1 (n) ≤
.
2
24(n + 1)
24n2
De Temple also stated without proof the following more accurate estimation:
1
Hn − γ = log(n + 21 ) +
− r2 (n),
24(n + 12 )2
where
(3)
7
7
≤ r2 (n) ≤
.
4
960(n + 1)
960n4
We will prove both these estimations. Without any more effort, the same method
applies to the more general sum
Hn (x) =
n
X
r=1
1
.
r+x
With this notation, Hn = Hn (0). We write C(x) = limn→∞ [Hn (x) − log n]. The existence
of the limit C(x) follows from elementary integral estimation of the sum Hn (x), but it will
also be established in the proof to follow. Clearly, C(0) = γ. We will establish a variant of
De Temple’s estimations applying to Hn (x) − C(x).
In fact, C(x) can be expressed in terms of the digamma function ψ(x) = Γ0 (x)/Γ(x).
From Euler’s limit formula for the gamma function, we have ψ(x + 1) = limn→∞ ψn (x + 1),
where
ψn (x + 1) = log n −
n
X
r=1
1
1
= log n − Hn (x).
x+r
So in fact C(x) = −ψ(x + 1). The resulting estimations will equate to a variant of Stirling’s
formula.
Note that
log(x +
1
)
2
− log(x −
1
)
2
Z
x+1/2
=
x−1/2
1
dt.
t
The mid-point approximation to this integral is 1/x. The proofs are based on estimates (at
various levels of accuracy) comparing log(x + 12 ) − log(x − 12 ) with 1/x.
PROPOSITION 1. For x > 21 ,
log(x + 21 ) − log(x − 21 ) =
where
1
+ δ1 (x),
x
(4)
1
1
≤ δ1 (x) ≤
.
3
12x
12(x − 21 )3
Proof. Let δ1 (x) be defined by (4). Then δ1 (x) → 0 as x → ∞, so δ1 (x) = −
R∞
x
δ10 (t)dt.
Now
1
1
1 −
t− 2
t+
1
1
= 2 1 − 2
t
t −4
1
=
2
2
4t (t − 14 )
1
>
.
4t4
−δ10 (t) =
Hence
Z
δ1 (x) >
x
2
2
Also, t (t −
1
)
4
> (t −
1 4
),
2
∞
1
2
−
1
t2
1
1
dt =
.
4
4t
12x3
so
Z
δ1 (x) <
x
∞
1
1
.
1 4 dt =
4(t − 2 )
12(x − 12 )3
THEOREM 1. Let n ≥ 1 and x > −1. Let C(x) = limn→∞ (Hn (x) − log n). Then
Hn (x) − C(x) = log(n + x + 12 ) + r1 (n, x),
where
(5)
1
1
≤ r1 (n, x) ≤
.
2
24(n + x + 1)
24(n + x)2
In particular,
Hn − γ = log(n + 21 ) + r1 (n),
2
(6)
where
1
1
≤
r
(n)
≤
1
24(n + 1)2
24n2
Also, (5) holds with n = 0 for x > 0 (with H0 (x) = 0), so that
ψ(x + 1) = log(x + 21 ) + r1 (x),
where
(7)
1
1
≤ r1 (x) ≤
.
2
24(x + 1)
24x2
Proof. Apply (4) to r + x for 1 ≤ r ≤ n and add: we obtain
log(n + x +
1
)
2
− log(x +
1
)
2
= Hn (x) +
n
X
δ1 (r + x),
r=1
equivalently
Hn (x) − log(n + x + 21 ) = − log(x + 12 ) −
n
X
δ1 (r + x).
(8)
r=1
Take the limit as n → ∞. Clearly, log(n + x +
C(x) = − log(x +
1
)
2
1
)
2
− log n → 0, so we obtain
−
∞
X
δ1 (r + x).
(9)
r=1
Taking the difference between (8) and (9), we obtain
Hn (x) − log(n + x +
1
)
2
− C(x) =
∞
X
δ1 (r + x).
(10)
r=n+1
Denote this by r1 (n, x). The condition x > −1 ensures that the inequality (4) applies to
δ1 (r + x) for r ≥ 2. By integral estimation, we now have for n ≥ 1,
∞
X
1
r1 (n, x) ≥
>
12(r + x)3
r=n+1
At the same time,
r1 (n, x) ≤
Z
∞
n+1
1
1
dt =
.
3
12(t + x)
24(n + x + 1)2
∞
X
1
.
12(r + x − 12 )3
r=n+1
The function 1/(t + x)3 is convex, and convex functions h(t) satisfy h(y − 12 ) ≤
hence
Z
r1 (n, x) ≤
n
∞
Ry
y−1
h(t) dt,
1
1
dt
=
.
12(t + x)3
24(n + x)2
For x > 0, the case n = 0 follows similarly from (9) (note that (4) now applies also to
δ1 (1 + x)). Since C(x) = −ψ(x + 1), this equates to (7).
3
Note. In (6) and (7), we have used the notation r1 (n) for r1 (n, 0) and r1 (x) for r1 (0, x).
The case x = − 12 gives the following estimation for sums of odd reciprocals.
COROLLARY 1.1. Let Un =
Pn
1
r=1 2r−1 .
Then
Un − 12 γ − log 2 = 12 log n + ρ1 (n),
where
(11)
1
1
.
1 2 ≤ ρ1 (n) ≤
48(n + 2 )
48(n − 12 )2
Proof. Note that 2Un = Hn (− 21 ). Also, 2Un = 2H2n − Hn ; it follows easily that
2Un − log n → γ + 2 log 2 as n → ∞. The statement is the case x = − 21 in Theorem 1.
Integration of (7) delivers the following variant of Stirling’s formula:
COROLLARY 1.2. For a certain constant c, we have
log Γ(x) = (x − 12 ) log(x − 21 ) − x + c − P1 (x),
where
(12)
1
1
≤ P1 (x) ≤
.
24x
24(x − 1)
Proof. Write (7) in the form ψ(t) = log(t − 12 ) + r1 (t − 1) and integrate on [ 23 , x] to
obtain
log Γ(x) −
Let I =
R∞
r (t
3/2 1
log Γ( 23 )
= (x −
1
) log(x
2
−
1
)
2
3
2
Z
x
−x+ +
r1 (t − 1) dt.
3/2
− 1) dt. Then (12) holds with c = log Γ( 23 ) + 32 + I and
Z
∞
Z
∞
r1 (t − 1) dt ≤
P1 (x) =
x
x
1
1
dt =
,
2
24(t − 1)
24(x − 1)
and similarly P1 (x) ≥ 1/24x.
As in most proofs of Stirling’s formula, one now needs to invoke the Wallis product (or
some equivalent) to evaluate c.
We now improve upon the estimate in Proposition 1, and derive the generalised verson
of (3).
PROPOSITION 2. For x > 12 ,
log(x +
1
)
2
− log(x −
1
)
2
1
1
= +
x 24
4
1
1
1 2 −
(x − 2 )
(x + 12 )2
− δ2 (x),
(13)
where
7
7
.
≤
δ
(x)
≤
2
240x5
240(x − 21 )5
Proof. Use (13) as the definition of δ2 (x). Using the expression for δ10 (t) from Proposition 1, we obtain
1
−δ20 (t) = −
1
)
4
+
4t2 (t2 −
G(t)
=
,
2
12t (t2 − 14 )3
1
1
1 3 −
12(t − 2 )
12(t + 12 )3
where
G(t) = −3(t2 − 41 )2 + t2 (t2 + 12 )3 − (t2 − 12 )3
= −3(t4 − 21 t2 +
=
7 2
t
4
=
7 2
(t
4
−
3
16
−
3
).
28
Hence
−δ20 (t) >
and
Z
7
48t2 (t2
∞
δ2 (x) = −
δ20 (t)
−
1 2
)
4
Z
∞
dt >
x
x
Also
−δ20 (t) <
+ t2 (3t2 + 14 )
1
)
16
48(t2
so that
δ2 (x) <
>
7
48t6
7
7
dt
=
.
48t6
240x5
7
7
,
1 3 <
− 4)
48(t − 12 )6
7
.
240(x − 12 )5
THEOREM 2. With the notation of Theorem 1, we have for n ≥ 1 and x > −1,
Hn (x) − C(x) = log(n + x + 21 ) +
where
(14)
7
7
≤
r
(n,
x)
≤
.
2
960(n + x + 1)4
960(n + x)4
In particular,
Hn − γ = log(n + 12 ) +
where
1
− r2 (n, x),
24(n + x + 21 )2
1
− r2 (n),
24(n + 12 )2
7
7
≤
r
(n)
≤
.
2
960(n + 1)4
960n4
5
(15)
Also, for x > 0,
ψ(x + 1) = log(x + 12 ) +
where
1
− r2 (x),
24(x + 12 )2
(16)
7
7
≤ r2 (x) ≤
.
4
960(x + 1)
960x4
Proof. Modify the proof of Theorem 1 as follows. Applying (13) to r + x for 1 ≤ r ≤ n
and adding, we obtain
n
Hn (x) − log(n + x +
1
)
2
= − log(x +
1
)
2
X
1
1
−
+
+
δ2 (r + x).
24(x + 12 )2 24(n + x + 12 )2 r=1
1
)
2
X
1
−
+
δ2 (r + x).
24(x + 12 )2 r=1
Taking the limit as n → ∞, we obtain
∞
C(x) = − log(x +
and taking the difference, we have
Hn (x) − log(n + x + 21 ) − C(x) =
where
r2 (n, x) =
∞
X
1
− r2 (n, x),
24(n + x + 12 )2
δ2 (r + x).
r=n+1
By (13) and integral estimation as before, we obtain the stated bounds for r2 (n, x).
COROLLARY 2.1. Let Un =
Pn
1
r=1 2r−1 .
Then
Un − 12 γ − log 2 = 12 log n +
where
1
− ρ2 (n),
48n2
7
7
.
1 4 ≤ ρ2 (n) ≤
1920(n + 2 )
1920(n − 12 )4
(17)
Note that (17) is considerably sharper than the estimate of the same type found by
applying (1) to H2n and Hn .
Reasoning as in Corollary 1.2, with (7) replaced by (16), we obtain:
COROLLARY 2.2. With c as in Corollary 1.2, we have
log Γ(x) = (x − 21 ) log(x − 21 ) − x + c −
6
1
+ P2 (x),
24(x − 12 )
(18)
where
7
7
≤
P
(x)
≤
.
2
2880x3
2880(x − 1)3
We derive a lower bound of a rather different type, which implies one given in [Ch].
For simplicity, we only state it for Hn instead of Hn (x).
COROLLARY 2.3. For all n ≥ 1,
Hn − γ ≥ log(n + 21 ) +
1
24(n + 12 +
1 2.
)
4n
(19)
Proof. By elementary algebra, for a, b > 0,
1
1
2b
−
>
.
a2 (a + b)2
(a + b)3
Hence for n ≥ 2 (with c ≤ 1 to be chosen),
1
2c
1
.
1 2 −
1
c 2 >
n(n + 1)3
(n + 2 )
(n + 2 + n )
To derive (19) from (15), we require this to be not less than 7/(40n4 ), which equates to
[(n + 1)/n]3 ≤
80
c.
7
With c = 14 , this holds for n ≥ 3, and one can check that (19) also holds
for n = 1 and n = 2.
Chen actually gives the lower bound 1/[24(n + a)2 ], where a ≈ 0.55106, and upper
bound 1/[24(n + 12 )2 ].
Negoi [Neg] demonstrated that the n−2 term in (15) can be absorbed into the log term
1
by considering log h(n), where h(n) = n + 21 + 24n
. His proof was by another variation of De
Temple’s method, entailing a rather laborious calculation. Here we show how to deduce his
result (in fact, a slightly stronger one) from Theorem 2.
COROLLARY 2.4. Let h(n) = n + 21 +
1
.
24n
Then
Hn − γ = log h(n) − ρ(n),
where
1
1
≤
ρ(n)
≤
.
48(n + 1)3
48(n + 12 )3
(Negoi’s statement has ρ(n) ≤ 1/(48n3 ).)
Proof. Write n +
1
2
= m. By Theorem 2,
Hn − γ ≤ log m +
7
1
24m2
(20)
1
(we do not need the term r2 (n)). Now m = h(n) − 24n
, so by the inequality log(1 − x) < −x,
we have
log m ≤ log h(n) −
1
.
24nh(n)
The stated upper bound in (20) will follow if we can show that
1
1
1
− 2 ≥
,
nh(n) m
2m3
which equates to 2m[m2 − nh(n)] ≥ nh(n). One checks that m2 − nh(n) = 12 n +
2m[m2 − nh(n)] = (n + 12 )(n +
5
)
12
5
,
24
so that
> nh(n).
The lower bound is proved in a similar way, using
log h(n) ≤ log m +
1
.
24mn
Of course, the term r2 (n) must be taken into account. We omit the details.
Comparison with
1
2
log(n2 + n + 31 )
In a development of Negoi’s result, Lu [Lu] has demonstrated that approximation to the
accuracy O(n−4 ) is obtained by comparing instead with
1
2
log f (n), where f (n) = n2 + n + 31 .
He showed that
Hn − γ = 12 log f (n) − r3 (n),
where n4 r3 (n) →
1
180
(21)
as n → ∞.
The relation of (21) to (15) is reflected by the fact that log(n + 21 ) = 21 log(n2 + n + 41 ).
Also, (21) explains the choice of the term
1
24n
in Negoi’s result (20), because this ensures
that h(n)2 = n2 + n + 13 + O( n1 ).
By a further application of De Temple’s method, we will present a generalisation of
Lu’s result, applying to Hn (x) − C(x) instead of Hn − γ, and incorporating explicit upper
and lower bounds.
PROPOSITION 3. Let f (x) = x2 + x + 13 . Then for all x ≥ 1,
log f (x) − log f (x − 1) =
where
2
− δ3 (x),
x
2
2
< δ3 (x) <
.
5
45x
45(x − 12 )5
8
(22)
Proof. We start with f (x) = x2 + x + c and allow the choice of c to emerge from the
reasoning. Let δ3 (x) be defined by (22). Now f (x)/f (x − 1) → 1, and hence δ3 (x) → 0, as
R∞
x → ∞. So δ3 (x) = − x δ30 (t) dt for all x > 0. Now
2t + 1
2t − 1
2
−
+ 2
f (t)
f (t − 1) t
G(t)
,
= 2
t f (t)f (t − 1)
−δ30 (t) =
where
G(t) = t2 (2t + 1)(t2 − t + c) − t2 (2t − 1)(t2 + t + c) + 2(t2 + t + c)(t2 − t + c)
= −t2 (2t2 − 2c) + 2[t4 + (2c − 1)t2 + c2 ]
= 2(3c − 1)t2 + 2c2 .
To eliminate the t2 term, we now choose c = 31 , so that G(t) =
−δ30 (t) =
Now f (t)f (t − 1) = t4 − 31 t2 +
1
9
2
9
and
2
.
9t2 f (t)f (t − 1)
< t4 for t ≥ 1, so
Z ∞
2
2
dt
=
δ3 (x) >
9t6
45x5
x
for x ≥ 1. On the other hand, f (t) > f (t − 1) > (t − 21 )2 , hence
Z ∞
2
2
δ3 (x) <
.
1 6 dt =
9(t − 2 )
45(x − 12 )5
x
THEOREM 3. Define Hn (x) and C(x) as in Theorem 1. Let f (t) = t2 + t + 13 . Then
for n ≥ 1 and x > −1,
Hn (x) − C(x) = 21 log f (n + x) − r3 (n, x),
where
(23)
1
1
≤ r3 (n, x) ≤
.
4
180(n + x + 1)
180(n + x)4
In particular,
Hn − γ = 21 log f (n) − r3 (n),
where
(24)
1
1
≤ r3 (n) ≤
.
4
180(n + 1)
180n4
Also, (23) holds with n = 0 for x > 0, so that
ψ(x + 1) = 21 log f (x) − r3 (x),
9
(25)
where
1
1
≤
r
(x)
≤
.
3
180(x + 1)4
180x4
Proof. Apply the identity (22) to r + x for 1 ≤ r ≤ n and add: we find
log f (n + x) − log f (x) = 2Hn (x) −
n
X
δ3 (r + x),
r=1
equivalently
2Hn (x) − log f (n + x) = − log f (x) +
n
X
δ3 (r + x).
(26)
r=1
Now f (n + x)/n2 → 1, so log f (n + x) − 2 log n → 0, as n → ∞. Taking the limit in (26),
we see that
2C(x) = − log f (x) +
∞
X
δ3 (r + x).
(27)
r=1
Now taking the difference, we have
2Hn (x) − log f (n + x) − 2C(x) = −2r3 (n, x),
where
∞
X
2r3 (n, x) =
(28)
δ3 (r + x).
r=n+1
If x > −1, then the inequality (22) applies to δ3 (r + x) for r ≥ 2. By integral estimation, we
now have for n ≥ 1,
∞
X
1
r3 (n, x) ≥
>
45(r + x)5
r=n+1
At the same time,
r3 (n, x) ≤
By convexity of 1/(t + x)5 , we deduce
Z ∞
r3 (n, x) ≤
n
Z
∞
1
1
dt =
.
5
45(t + x)
180(n + x + 1)4
n+1
∞
X
1
.
45(r − 12 + x)5
r=n+1
1
1
dt =
.
5
45(t + x)
180(n + x)4
For x > 0, the case n = 0 follows similarly from (27): this equates to (25).
Note 1. The upper bounds for δ3 (x) in Proposition 3 and r3 (n, x) in Theorem 1 can be
slightly improved. One can verify that t2 f (t)f (t − 1) ≥ (t −
(t − 12 )6 . This leads to δ3 (x) < 2/[45(x −
1 5
)]
12
1 6
)
12
where we previously used
and r3 (n, x) ≤ 1/[180(n + x +
10
5 4
) ].
12
1
−5
So 12 log f (n) − 180n
). With n chosen to be only 10,
4 approximates Hn with error O(n
the resulting lower and upper estimates for γ are 0.57721559 and 0.57721577 to eight d.p.
(compare the actual value 0.57721566.)
Again, the case x = − 21 gives an estimation for sums of odd reciprocals.
COROLLARY 3.1. Let Un =
Pn
1
r=1 2r−1 .
Then
Un − 12 γ − log 2 = 14 log(n2 +
where
1
)
12
− ρ3 (n),
(29)
1
1
.
1 4 ≤ ρ3 (n) ≤
360(n + 2 )
360(n − 12 )4
Proof. Just note that f (n − 12 ) = n2 +
1
.
12
A Stirling-type formula derived from Theorem 3 would involve the rather unpleasant
antiderivative of log f (x).
The estimate for Hn − γ in Theorem 3 has a rather surprising application to a residual
term that arises in the Dirichlet divisor problem. See [Jam1].
Further note on the selection of c. The choice c =
1
3
in Proposition 3 emerged from the
proof. We now show directly how this choice in Theorem 3 can be determined ahead of the
actual proof. Using (1), we show that if f (n) = n2 +n+c and Hn −γ = 12 log f (n)+O(1/n3 ),
then c = 31 . By (1),
1
1
1
2H(n) − 2γ = 2 log n + − 2 + O 4 ,
n 6n
n
so
1
1
1
log f (n) − 2 log n = − 2 + O 3 .
n 6n
n
But by the logarithmic series,
1
c
log f (n) − 2 log n = log 1 + + 2
n n
2
1
c
1 1
c
1
=
+ 2−
+ 2 +O 3
n n
2 n n
n
1 c − 12
1
=
+O 3 .
+
2
n
n
n
Hence c −
1
2
= − 16 , so c = 31 .
Comparison of Theorems 2 and 3. The estimates in Theorems 2 and 3 are clearly
of similar degrees of accuracy. We compare them, using the logarithmic series. For this
11
purpose, write n + x = y and
log(y + 12 ) +
1
= L(y).
24(y + 12 )2
We restate Theorem 2 in the form
L(y) −
7
7
≤ Hn (x) − C(x) ≤ L(y) −
4
960y
960(y + 1)4
(30)
and Theorem 3 in the form
1
2
log f (y) −
Now f (y) = (y + 12 )2 +
1
,
12
1
1
≤ Hn (x) − C(x) ≤ 12 log f (y) −
.
4
180y
180(y + 1)4
(31)
so
log f (y) = 2 log(y +
1
)
2
+ log 1 +
1
12(y + 12 )2
.
Since t − 21 t2 ≤ log(1 + t) ≤ t for 0 < t < 1, we have
L(y) −
Given (31), apply (32) with y +
L(y) −
1
≤ 21 log f (y) ≤ L(y).
576(y + 12 )4
1
2
replaced by y. Noting that
1
180
+
(32)
1
576
=
7
,
960
we obtain
7
1
≤ Hn (x) − C(x) ≤ L(y) −
,
4
960y
180(y + 1)4
in which the lower bound agrees with (30), while the upper bound is slightly weaker. In fact,
the upper bound stated in (30) can be derived from (31) using the inequality log(1 + t) ≤
t − 21 t2 + 31 t3 , after some manipulation.
Conversely, (30) implies
1
2
log f (y) −
7
1
7
≤ Hn (x) − C(x) ≤ 12 log f (y) +
.
1 4 −
4
960y
960(y + 1)4
576(y + 2 )
Both bounds are weaker than those in (31), though the upper bound differs by no more than
1/(288y 5 ).
Another method: mid-point Euler-Maclaurin summation
In [DTW], a mid-point variant of Euler-Maclaurin summation is used to derive an
asymptotic expansion for Hn − γ in terms of (n + 21 )−2k . The coefficients are in terms
of the Bernoulli numbers Bn ( 12 ). The method is described in [DTW] for the special case
f (t) = 1/t, and for general functions in my website notes [Jam2]. Taken as far as the third
derivative, and expressed as a pair of inequalities, it can be stated as follows. Suppose that
12
f is completely monotonic on (0, ∞), that is, f (2k) (t) ≥ 0 and f (2k+1) (t) ≤ 0 for all k ≥ 0
and t > 0. Then
n
X
Z
n+ 12
f (r) =
m− 21
r=m
1 0
0
1
1
f (n + 2 ) − f (m − 2 ) + T2 (m, n),
f (t) dt −
24
where
7 (3)
f (n + 21 ) − f (3) (m − 12 ) .
5760
Applied with f (t) = 1/(t + x), this leads to the following stronger upper bound for r2 (n, x)
0 ≤ T2 (m, n) ≤
in Theorem 2:
r2 (n, x) ≤
7
.
960(n + x + 12 )4
Further terms of the expansion give stronger bounds (both upper and lower), at the cost of
greater complication.
References
[BM]
George Boros and Victor Moll, Irresistible Integrals, Cambridge Univ. Press
(2004).
[Ch]
Chao-Ping Chen, Inequalities for the Euler-Mascheroni constant, Appl. Math.
Lett. 23 (2010), 161–164.
[DT]
D. W. De Temple, A quicker convergence to Euler’s constant, American Math.
Monthly 100 (1993), 468–470.
[DTW]
D. W. De Temple and S. H. Wang, Half integer approximations for the partial
sums of the harmonic series, J. Math. Anal. Appl. 160 (1991), 187–190.
[Lu]
Dawei Lu, Some quicker classes of sequences convergent to Euler’s constant,
Appl. Math. Comp. 232 (2014), 172–177.
[Jam1]
A sharp estimate for harmonic sums and the digamma function, with an application to the Dirichlet divisor problem, preprint.
[Jam2]
Euler-Maclaurin summation, at www.maths.lancs.ac.uk/˜jameson.
[Neg]
Tanase Negoi, A faster convergence to Euler’s constant, Math. Gazette 83
(1999), 487–489.
updated 29 January 2016
13