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Journal of Inequalities and Applications

Erschienen in:

Open Access 01.12.2013 | Research

On the harmonic number expansion by Ramanujan

verfasst von: Cristinel Mortici, Chao-Ping Chen

Erschienen in: Journal of Inequalities and Applications | Ausgabe 1/2013

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Abstract

Let

γ = 0.577215664 \dots

denote the Euler-Mascheroni constant, and let the sequences

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-222/MediaObjects/13660_2012_Article_679_Equa_HTML.gif

The main aim of this paper is to find the values r, s, t, a, b, c and d which provide the fastest sequences

{(u_{n})}_{n \geq 1}

and

{(v_{n})}_{n \geq 1}

approximating the Euler-Mascheroni constant. Also, we give the upper and lower bounds for

\sum_{k = 1}^{n} \frac{1}{k} - \frac{1}{2} ln (n^{2} + n + \frac{1}{3}) - γ

in terms of

n^{2} + n + \frac{1}{3}

MSC: 11Y60, 40A05, 33B15.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

CM proposed the sequence

u_{n}

. CPC proposed the sequence

v_{n}

. CM proposed to solve the problems using Lemma 1, while CPC used Lemma 2 in evaluations. Both authors made the computations and verified their corectedness. The authors read and approved the final manuscript.

1 Introduction

The Euler-Mascheroni constant

γ = 0.577215664 \dots

is defined as the limit of the sequence

D_{n} = H_{n} - ln n,

(1.1)

where

H_{n}

denotes the n th harmonic number defined for

n \in N : = {1, 2, 3, \dots}

H_{n} = \sum_{k = 1}^{n} \frac{1}{k} .

Several bounds for

D_{n} - γ

have been given in the literature [1‐7]. For example, the following bounds for

D_{n} - γ

were established in [3, 7]:

\frac{1}{2 (n + 1)} < D_{n} - γ < \frac{1}{2 n} (n \in N) .

The convergence of the sequence

D_{n}

to γ is very slow. Some quicker approximations to the Euler-Mascheroni constant were established in [8‐21]. For example, Cesàro [8] proved that for every positive integer

n \geq 1

, there exists a number

c_{n} \in (0, 1)

such that the following approximation is valid:

\sum_{k = 1}^{n} \frac{1}{k} - \frac{1}{2} ln (n^{2} + n) - γ = \frac{c_{n}}{6 n (n + 1)} .

Entry 9 of Chapter 38 of Berndt’s edition of Ramanujan’s Notebooks [[22], p.521] reads,

‘Let

m : = \frac{n (n + 1)}{2}

, where n is a positive integer. Then, as n approaches infinity,

\begin{array}{rcl} \sum_{k = 1}^{\infty} \frac{1}{k} & \sim & \frac{1}{2} ln (2 m) + γ + \frac{1}{12 m} - \frac{1}{120 m^{2}} + \frac{1}{630 m^{3}} - \frac{1}{1, 680 m^{4}} + \frac{1}{2, 310 m^{5}} \\ - \frac{191}{360, 360 m^{6}} + \frac{1}{30, 030 m^{7}} - \frac{2, 833}{1, 166, 880 m^{8}} + \frac{140, 051}{17, 459, 442 m^{9}} - [\dots] . ’ \end{array}

For the history and the development of Ramanujan’s formula, see [20].

Recently, by changing the logarithmic term in (1.1), DeTemple [15], Negoi [18] and Chen et al. [14] have presented, respectively, faster and faster asymptotic formulas as follows:

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-222/MediaObjects/13660_2012_Article_679_Equ2_HTML.gif

(1.2)

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-222/MediaObjects/13660_2012_Article_679_Equ3_HTML.gif

(1.3)

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-222/MediaObjects/13660_2012_Article_679_Equ4_HTML.gif

(1.4)

Chen and Mortici [13] provided a faster asymptotic formula than those in (1.2) to (1.4),

\sum_{k = 1}^{n} \frac{1}{k} - ln (n + \frac{1}{2} + \frac{1}{24 n} - \frac{1}{48 n^{2}} + \frac{23}{5, 760 n^{3}}) = γ + O (n^{- 5}) (n \to \infty),

(1.5)

and posed the following natural question.

Open problem For a given positive integer p, find the constants

a_{j}

(

j = 0, 1, 2, \dots, p

) such that

\sum_{k = 1}^{n} \frac{1}{k} - ln (n + \sum_{j = 0}^{p} \frac{a_{j}}{n^{j}})

(1.6)

is the sequence which would converge to γ in the fastest way.

Very recently, Yang [21] published the solution of the open problem (1.6) by using logarithmic-type Bell polynomials.

For all

n \in N

, let

P_{n} = \sum_{k = 1}^{n} \frac{1}{k} - \frac{1}{2} ln (n^{2} + n + \frac{1}{3})

(1.7)

and

Q_{n} = \sum_{k = 1}^{n} \frac{1}{k} - \frac{1}{4} ln [{(n^{2} + n + \frac{1}{3})}^{2} - \frac{1}{45}] .

Chen and Li [12] proved that for all integers

n \geq 1

\frac{1}{180 {(n + 1)}^{4}} < γ - P_{n} < \frac{1}{180 n^{4}}

(1.8)

and

\frac{8}{2, 835 {(n + 1)}^{6}} < Q_{n} - γ < \frac{8}{2, 835 n^{6}} .

Now we define the sequences

u_{n} = \sum_{k = 1}^{n} \frac{1}{k} - \frac{1}{2} ln (n^{2} + n + \frac{1}{3}) - \frac{1}{r (n^{2} + n + \frac{1}{3}) + s {(n^{2} + n + \frac{1}{3})}^{2} + t}

(1.9)

and

\begin{array}{rcl} v_{n} & = & \sum_{k = 1}^{n} \frac{1}{k} - \frac{1}{2} ln (n^{2} + n + \frac{1}{3}) \\ - (\frac{a}{{(n^{2} + n + \frac{1}{3})}^{2}} + \frac{b}{{(n^{2} + n + \frac{1}{3})}^{3}} + \frac{c}{{(n^{2} + n + \frac{1}{3})}^{4}} + \frac{d}{{(n^{2} + n + \frac{1}{3})}^{5}}), \end{array}

(1.10)

respectively. Our Theorems 1 and 2 are to find the values r, s, t, a, b, c and d which provide the fastest sequences

{(u_{n})}_{n \geq 1}

and

{(v_{n})}_{n \geq 1}

approximating the Euler-Mascheroni constant.

Theorem 1 Let

{(u_{n})}_{n \geq 1}

be defined by (1.9). For

r = - \frac{640}{7}, s = - 180, t = \frac{26, 770}{441},

we have

lim_{n \to \infty} n^{11} (u_{n} - u_{n + 1}) = \frac{457, 528}{123, 773, 265}

(1.11)

and

lim_{n \to \infty} n^{10} (u_{n} - γ) = \frac{457, 528}{123, 773, 265} .

(1.12)

The speed of convergence of the sequence

{(u_{n})}_{n \geq 1}

n^{- 10}

Theorem 2 Let

{(v_{n})}_{n \geq 1}

be defined by (1.10). For

a = - \frac{1}{180}, b = \frac{8}{2, 835}, c = - \frac{5}{1, 512}, d = \frac{592}{93, 555},

we have

lim_{n \to \infty} n^{13} (v_{n} - v_{n + 1}) = - \frac{796, 801}{3, 648, 645} and lim_{n \to \infty} n^{12} (v_{n} - γ) = - \frac{796, 801}{43, 783, 740} .

The speed of convergence of the sequence

{(v_{n})}_{n \geq 1}

n^{- 12}

Our Theorems 3 and 4 establish the bounds for

γ - P_{n}

in terms of

n^{2} + n + \frac{1}{3}

Theorem 3 Let

P_{n}

be defined by (1.7). Then

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-222/MediaObjects/13660_2012_Article_679_Equ13_HTML.gif

(1.13)

Theorem 4 Let

P_{n}

be defined by (1.7). Then

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-222/MediaObjects/13660_2012_Article_679_Equ14_HTML.gif

(1.14)

Remark 1 The inequality (1.14) is sharper than (1.8), while the inequality (1.13) is sharper than (1.14).

2 Lemmas

Before we prove the main theorems, let us give some preliminary results.

The constant γ is deeply related to the gamma function

Γ (z)

thanks to the Weierstrass formula:

Γ (z) = \frac{e^{- γ z}}{z} \prod_{k = 1}^{\infty} {{(1 + \frac{z}{k})}^{- 1} e^{z / k}} (z \in C ∖ Z_{0}^{-}; Z_{0}^{-} : = {- 1, - 2, - 3, \dots}) .

The logarithmic derivative of the gamma function

ψ (z) = \frac{Γ^{'} (z)}{Γ (z)} or ln Γ (z) = \int_{1}^{z} ψ (t) d t

is known as the psi (or digamma) function. The successive derivatives of the psi function

ψ (z)

ψ^{(n)} (z) : = \frac{d^{n}}{d z^{n}} {ψ (z)} (n \in N)

are called the polygamma functions.

The following recurrence and asymptotic formulas are well known for the psi function:

ψ (z + 1) = ψ (z) + \frac{1}{z}

(2.1)

(see [[23], p.258]), and

ψ (z) \sim ln z - \frac{1}{2 z} - \frac{1}{12 z^{2}} + \frac{1}{120 z^{4}} - \frac{1}{252 z^{6}} + \dots (z \to \infty in | arg z | < π)

(2.2)

(see [[23], p.259]). From (2.1) and (2.2), we get

ψ (n + 1) \sim ln n + \frac{1}{2 n} - \frac{1}{12 n^{2}} + \frac{1}{120 n^{4}} - \frac{1}{252 n^{6}} + \dots (n \to \infty) .

(2.3)

It is also known [[23], p.258] that

ψ (n + 1) = - γ + \sum_{k = 1}^{n} \frac{1}{k} .

Lemma 1 [24, 25]

{(λ_{n})}_{n \geq 1}

is convergent to zero and there exists the limit

lim_{n \to \infty} n^{k} (λ_{n} - λ_{n + 1}) = l \in R,

with

k > 1

, then there exists the limit

lim_{n \to \infty} n^{k - 1} λ_{n} = \frac{l}{k - 1} .

Lemma 1 gives a method for measuring the speed of convergence.

Lemma 2 [[26], Theorem 9]

Let

k \geq 1

and

n \geq 0

be integers. Then, for all real numbers

x > 0

S_{k} (2 n; x) < {(- 1)}^{k + 1} ψ^{(k)} (x) < S_{k} (2 n + 1; x),

(2.4)

where

S_{k} (p; x) = \frac{(k - 1)!}{x^{k}} + \frac{k!}{2 x^{k + 1}} + \sum_{i = 1}^{p} [B_{2 i} \prod_{j = 1}^{k - 1} (2 i + j)] \frac{1}{x^{2 i + k}},

and

B_{i}

(

i = 0, 1, 2, \dots

) are Bernoulli numbers defined by

\frac{t}{e^{t} - 1} = \sum_{i = 0}^{\infty} B_{i} \frac{t^{i}}{i!} .

It follows from (2.4) that for

x > 0

\begin{aligned} \frac{1}{x} + \frac{1}{2 x^{2}} + \frac{1}{6 x^{3}} - \frac{1}{30 x^{5}} + \frac{1}{42 x^{7}} - \frac{1}{30 x^{9}} + \frac{5}{66 x^{11}} - \frac{691}{2, 730 x^{13}} \\ < ψ^{'} (x) < \frac{1}{x} + \frac{1}{2 x^{2}} + \frac{1}{6 x^{3}} - \frac{1}{30 x^{5}} + \frac{1}{42 x^{7}} - \frac{1}{30 x^{9}} + \frac{5}{66 x^{11}} - \frac{691}{2, 730 x^{13}} + \frac{7}{6 x^{15}}, \end{aligned}

from which we imply that for

x > 0

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-222/MediaObjects/13660_2012_Article_679_Equ19_HTML.gif

(2.5)

3 Proofs of Theorems 1-4

Proof of Theorem 1 By using the Maple software, we write the difference

u_{n} - u_{n + 1}

as a power series in

n^{- 1}

\begin{array}{rcl} u_{n} - u_{n + 1} & = & (- \frac{s + 180}{45 s}) \frac{1}{n^{5}} + (\frac{s + 180}{9 s}) \frac{1}{n^{6}} \\ + (\frac{2 (- 6, 048 s + 567 r - 32 s^{2})}{189 s^{2}}) \frac{1}{n^{7}} \\ + (\frac{2 (- 567 r + 2, 268 s + 11 s^{2})}{27 s^{2}}) \frac{1}{n^{8}} \\ + (\frac{2 (- 23 s^{3} + 2, 430 s r - 5, 310 s^{2} + 108 s t - 108 r^{2})}{27 s^{3}}) \frac{1}{n^{9}} \\ + (\frac{2 (- 13, 770 s r + 19, 170 s^{2} - 1, 620 s t + 1, 620 r^{2} + 73 s^{3})}{45 s^{3}}) \frac{1}{n^{10}} \\ + \frac{1}{2, 673 s^{4}} (- 15, 443 s^{4} + 4, 834, 566 s^{2} r - 4, 650, 624 s^{3} + 1, 033, 560 s^{2} t \\ - 1, 033, 560 s r^{2} - 53, 460 s r t + 26, 730 r^{3}) \frac{1}{n^{11}} \\ + O (\frac{1}{n^{12}}) . \end{array}

(3.1)

According to Lemma 1, we have three parameters r, s and t which produce the fastest convergence of the sequence from (3.1)

{\begin{matrix} s + 180 = 0, \\ - 6, 048 s + 567 r - 32 s^{2} = 0, \\ - 23 s^{3} + 2, 430 s r - 5, 310 s^{2} + 108 s t - 108 r^{2} = 0, \end{matrix}

namely if

r = - \frac{640}{7}, s = - 180, t = \frac{26, 770}{441} .

Thus, we have

u_{n} - u_{n + 1} = \frac{457, 528}{123, 773, 265 n^{11}} + O (\frac{1}{n^{12}}) .

By using Lemma 1, we obtain the assertion of Theorem 1. □

Proof of Theorem 2 By using the Maple software, we write the difference

v_{n} - v_{n + 1}

as a power series in

n^{- 1}

\begin{array}{rcl} v_{n} - v_{n + 1} & = & (- \frac{1}{45} - 4 a) \frac{1}{n^{5}} + (\frac{1}{9} + 20 a) \frac{1}{n^{6}} + (- 64 a - 6 b - \frac{64}{189}) \frac{1}{n^{7}} \\ + (\frac{22}{27} + 168 a + 42 b) \frac{1}{n^{8}} + (- \frac{1, 180}{3} a - 8 c - \frac{46}{27} - 180 b) \frac{1}{n^{9}} \\ + (72 c + \frac{146}{45} + 852 a + 612 b) \frac{1}{n^{10}} \\ + (- \frac{1, 160}{3} c - \frac{15, 443}{2, 673} - \frac{5, 426}{3} b - 10 d - \frac{46, 976}{27} a) \frac{1}{n^{11}} \\ + (\frac{2, 375}{243} + \frac{14, 542}{3} b + \frac{4, 840}{3} c + \frac{91, 432}{27} a + 110 d) \frac{1}{n^{12}} + O (\frac{1}{n^{13}}) . \end{array}

(3.2)

According to Lemma 1, we have four parameters a, b, c and d which produce the fastest convergence of the sequence from (3.2)

{\begin{matrix} - \frac{1}{45} - 4 a = 0, \\ - 64 a - 6 b - \frac{64}{189} = 0, \\ - \frac{1, 180}{3} a - 8 c - \frac{46}{27} - 180 b = 0, \\ - \frac{1, 160}{3} c - \frac{15, 443}{2, 673} - \frac{5, 426}{3} b - 10 d - \frac{46, 976}{27} a = 0, \end{matrix}

namely if

a = - \frac{1}{180}, b = \frac{8}{2, 835}, c = - \frac{5}{1, 512}, d = \frac{592}{93, 555} .

Thus, we have

v_{n} - v_{n + 1} = - \frac{796, 801}{3, 648, 645 n^{13}} + O (\frac{1}{n^{14}}) .

By using Lemma 1, we obtain the assertion of Theorem 2. □

Proof of Theorem 3 Here we only prove the second inequality in (1.13). The proof of the first inequality in (1.13) is similar. The upper bound of (1.13) is obtained by considering the function F for

x \geq 1

defined by

F (x) = \frac{1}{2} ln (x^{2} + x + \frac{1}{3}) - ψ (x + 1) - \frac{1}{\frac{640}{7} (n^{2} + n + \frac{1}{3}) + 180 {(n^{2} + n + \frac{1}{3})}^{2} - \frac{26, 770}{441}} .

Differentiation and applying the right-hand inequality of (2.5) yield

\begin{array}{rcl} F^{'} (x) & = & - ψ^{'} (x + 1) + \frac{2 x + 1}{2 (x^{2} + x + \frac{1}{3})} \\ + \frac{55, 566 (126 x^{3} + 189 x^{2} + 137 x + 37)}{5 {(7, 938 x^{4} + 15, 876 x^{3} + 17, 262 x^{2} + 9, 324 x - 451)}^{2}} \\ > & - (\frac{1}{x} - \frac{1}{2 x^{2}} + \frac{1}{6 x^{3}} - \frac{1}{30 x^{5}} + \frac{1}{42 x^{7}} - \frac{1}{30 x^{9}} + \frac{5}{66 x^{11}} - \frac{691}{2, 730 x^{13}} + \frac{7}{6 x^{15}}) \\ + \frac{2 x + 1}{2 (x^{2} + x + \frac{1}{3})} + \frac{55, 566 (126 x^{3} + 189 x^{2} + 137 x + 37)}{5 {(7, 938 x^{4} + 15, 876 x^{3} + 17, 262 x^{2} + 9, 324 x - 451)}^{2}} \\ = & \frac{P (x)}{30, 030 x^{13} {(3 x^{2} + 3 x + 1)}^{6}}, \end{array}

where

\begin{array}{rcl} P (x) & = & 35, 471, 898, 974, 548, 627, 145 + 138, 773, 138, 144, 376, 345, 519 (x - 4) \\ + 241, 909, 257, 272, 859, 643, 240 {(x - 4)}^{2} \\ + 253, 899, 751, 881, 744, 791, 655 {(x - 4)}^{3} \\ + 181, 059, 030, 163, 487, 870, 836 {(x - 4)}^{4} \\ + 93, 303, 260, 620, 236, 720, 571 {(x - 4)}^{5} \\ + 35, 932, 291, 146, 874, 735, 228 {(x - 4)}^{6} + 10, 519, 794, 292, 714, 982, 599 {(x - 4)}^{7} \\ + 2, 353, 926, 972, 956, 528, 576 {(x - 4)}^{8} + 400, 626, 844, 002, 342, 775 {(x - 4)}^{9} \\ + 51, 041, 813, 866, 867, 916 {(x - 4)}^{10} + 4, 719, 218, 347, 433, 667 {(x - 4)}^{11} \\ + 299, 247, 577, 164, 158 {(x - 4)}^{12} + 11, 646, 155, 626, 560 {(x - 4)}^{13} \\ + 209, 840, 641, 920 {(x - 4)}^{14} > 0 for x \geq 4 . \end{array}

Therefore,

F^{'} (x) > 0

for

x \geq 4

For

x = 1, 2, 3, 4

, we compute directly:

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-222/MediaObjects/13660_2012_Article_679_Equac_HTML.gif

Hence, the sequence

{(F (n))}_{n \geq 1}

is strictly increasing. This leads to

F (n) < lim_{n \to \infty} F (n) = 0

by using the asymptotic formula (2.3). This completes the proof of the second inequality in (1.13). □

Proof of Theorem 4 Here we only prove the first inequality in (1.14). The proof of the second inequality in (1.14) is similar. The lower bound of (1.14) is obtained by considering the function G for

x \geq 1

defined by

\begin{array}{rcl} G (x) & = & ψ (x + 1) - \frac{1}{2} ln (x^{2} + x + \frac{1}{3}) \\ + (\frac{\frac{1}{180}}{{(x^{2} + x + \frac{1}{3})}^{2}} + \frac{- \frac{8}{2, 835}}{{(x^{2} + x + \frac{1}{3})}^{3}} + \frac{\frac{5}{1, 512}}{{(x^{2} + x + \frac{1}{3})}^{4}} + \frac{- \frac{592}{93, 555}}{{(x^{2} + x + \frac{1}{3})}^{5}}) . \end{array}

Differentiation and applying the left-hand inequality of (2.5) yield

\begin{array}{rcl} G^{'} (x) & = & ψ^{'} (x + 1) - \frac{2 x + 1}{2 (x^{2} + x + \frac{1}{3})} \\ - \frac{3 (4, 158 x^{7} + 14, 553 x^{6} + 19, 701 x^{5} + 12, 870 x^{4} + 8, 283 x^{3} + 6, 831 x^{2} - 8, 276 x - 5, 194)}{770 {(3 x^{2} + 3 x + 1)}^{6}} \\ > & (\frac{1}{x} - \frac{1}{2 x^{2}} + \frac{1}{6 x^{3}} - \frac{1}{30 x^{5}} + \frac{1}{42 x^{7}} - \frac{1}{30 x^{9}} + \frac{5}{66 x^{11}} - \frac{691}{2, 730 x^{13}}) - \frac{2 x + 1}{2 (x^{2} + x + \frac{1}{3})} \\ - \frac{3 (41, 58 x^{7} + 14, 553 x^{6} + 19, 701 x^{5} + 12, 870 x^{4} + 8, 283 x^{3} + 6, 831 x^{2} - 8, 276 x - 5, 194)}{770 {(3 x^{2} + 3 x + 1)}^{6}} \\ = & \frac{Q (x)}{30, 030 x^{13} {(3 x^{2} + 3 x + 1)}^{6}}, \end{array}

where

\begin{array}{rcl} Q (x) & = & 274, 317, 996, 839, 484 + 1, 074, 684, 262, 984, 527 (x - 5) \\ + 1, 571, 352, 927, 565, 772 {(x - 5)}^{2} + 1, 266, 557, 271, 610, 345 {(x - 5)}^{3} \\ + 652, 427, 951, 634, 329 {(x - 5)}^{4} + 230, 639, 944, 842, 034 {(x - 5)}^{5} \\ + 57, 987, 546, 990, 473 {(x - 5)}^{6} + 10, 515, 845, 175, 406 {(x - 5)}^{7} \\ + 1, 371, 027, 303, 124 {(x - 5)}^{8} + 125, 702, 024, 549 {(x - 5)}^{9} \\ + 7, 709, 579, 845 {(x - 5)}^{10} + 284, 457, 957 {(x - 5)}^{11} \\ + 4, 780, 806 {(x - 5)}^{12} > 0 for x \geq 5 . \end{array}

Therefore,

G^{'} (x) > 0

for

x \geq 5

For

x = 1, 2, 3, 4, 5

, we compute directly:

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-222/MediaObjects/13660_2012_Article_679_Equah_HTML.gif

Hence, the sequence

{(G (n))}_{n \geq 1}

is strictly increasing. This leads to

G (n) < lim_{n \to \infty} G (n) = 0

by using the asymptotic formula (2.3). This completes the proof of the first inequality in (1.14). □

Remark 2 Some calculations in this work were performed by using the Maple software for symbolic calculations.

Remark 3 The work of the first author was supported by a grant of the Romanian National Authority for Scientific Research, CNCS-UEFISCDI, project number PN-II-ID-PCE-2011-3-0087.

Acknowledgements

Dedicated to Professor Hari M Srivastava.

Open AccessThis article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

CM proposed the sequence

u_{n}

. CPC proposed the sequence

v_{n}

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Titel: On the harmonic number expansion by Ramanujan
verfasst von: Cristinel Mortici
Chao-Ping Chen
Publikationsdatum: 01.12.2013
Verlag: Springer International Publishing
Erschienen in: Journal of Inequalities and Applications / Ausgabe 1/2013
Elektronische ISSN: 1029-242X
DOI: https://doi.org/10.1186/1029-242X-2013-222

Springer Professional

On the harmonic number expansion by Ramanujan

Abstract

Competing interests

Authors’ contributions

1 Introduction

2 Lemmas

3 Proofs of Theorems 1-4

Acknowledgements

Competing interests

Authors’ contributions

Premium Partner

Springer Professional

Abstract

Competing interests

Authors’ contributions

1 Introduction

2 Lemmas

3 Proofs of Theorems 1-4

Acknowledgements

Competing interests

Authors’ contributions

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