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Boris Gourévitch
The world of Pi - V2.57
modif. 13/04/2013



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13 Harmonic Series

The BBP formulae are expressed for example as linear combination of those hypergeometric functions but even more things await us! Because thanks to integral representation, we can also in fact obtain harmonic series of the same form as all those that we just found.

The harmonic series are those which the terms contain the harmonic sum         sum 
H(n) =   nk=1 1k  .

13.1 Links between harmonic series and the polylogarithms

We have seen that the BBP or factorial formulae are integral combinations of the kind  integral  ulnk(y)R(y)
   ----------dy
 0    Q (y)  .

Let us now consider the integral equivalent to the serie's term of the harmonic sum  sum o o  H(n)xn
  n=1  . According to the product formula by Cauchy for which two absolutly convergant series  sum  a
    n  and  sum  b
   n

 oo  sum       ( sum  oo  ) (  oo  sum    )
   wn =     an       bn
n=0      n=0      n=0
(587)

where       sum n
wn =   k=0akbn-k  , by choosing       k
ak = xk-  and bk = xk  we get wn = xnH(n)  and so

  oo          (  oo     )(   oo   )
 sum  H(n)xn =    sum  xn-    sum  xn  = - ln-(1--x)
n=0           n=0 n    n=0          1- x
(588)

You can start to see what I mean... :-) As soon as an integral uses this kind of formula, we will have some harmonic lurking in a corner. Here, we find one immediatly in x = 12  the formula

+ sum  oo -Hn-
   2n+1 = ln (2)
n=1
(589)

With ln(1-x)
  x  we had some polylogarithm (voir 35), but this doesn't matter, we divide the expression 588 by x ! And then integrating between 0  and 1  , we have

pict

  But, since L2 (1)= p2 - 1ln2(2)
    2   12   2  , we finaly obtain that

+ sum  oo  Hn    p2
    n2n = 12-
n=1
(591)

Impressive, no? obviously, this remind us of many things, we are swimming in a sea of polylogarithm and logarithm...

We can study in more generality the harmonic series of this kind and the formulae that follows from them by introducing the notations by Gery Huvent (him again !). All the following formulae which are not yet known can be credited to him !

13.2 Study of        +o sum  o  Hk
fkp(x) =    nnp xn
       n=1  and of        + sum  oo   Hk
gkp(x) =   (n+n1)pxn+1
       n=1

13.2.1 Definition, remarquable relations

We let

pict
pict

Those series have a convergence radius of 1. The convergence take place on the border as soon as p > 1  (because Hn  ~ ln(n)  ). Then

pict

Proof.

pict

hence

pict

Similarly

          (       -1-)k    n           k-i
--Hkn---=  -Hn+1---n+1---=  sum  (- 1)iCi--Hn+1---
(n + 1)p      (n+ 1)p      i=0       k(n + 1)p+i
(605)

 _

13.2.2 Calculations of certain functions

A usual expansion (donne by Cauchy's product) gives

  1       ln(1--x)-
f0 (x) = -  1 - x
(606)

Then by integrations

 1     1   2                1     2
f1 (x) = 2 ln (1 - x)+ L2(x) = 2L1(x) + L2 (x)
(607)

and

         2
f12 (x) = ln-(1--x)-ln(x)-+ ln(1 -x) L2(1- x)+ L3 (x) - L3(1- x)+ z (3)
              2
(608)

Before carrying on, let us examine this equality. For x = - 1,  the serie defining f12  converges, but the right hand side of the previous equality does not. However with the formulae by Euler and Landen, we get

pict

The last equality gives us an expression of f21(x)  valid for [-1, 1]
    2 . By analytical expansion, we know that the different expression obtained coincide where they are defined simultaneously.
We will hence use if necessairy different expressions for the calculations of functions fpk  in x = -1  and x = 1
    2  .

pict

We deduce from the calculation of f11  that

 2      ln2(1--x)-+-L2(x)
f0 (x) =      1- x
(611)

then by integration that

pict

Finaly

          2
 3      -p2-ln-(1---x)+-32 ln2(1--x)ln(x)--ln3(1--x)+-L3(x)+-3L3-(1--x)--3z(3)
f0 (x) =                               1- x
(613)

and hence

We deduce from it that

pict

pict
pict

Then by a differentiation calculation, we have

pict

which gives for real x

pict

and

pict

and finally

pict

13.3 Applications to the calculation of certain series

We use the previous results with some rightly chosen values for x

13.3.1 With x = 1

The values are more easily obtained with the Beta functions, for example

+o o       4
 sum   Hn-= p--
k=1 n3   72
(628)

13.3.2 With x = 1
    2

We obtain by taking the value of the functions fk
 p  and gk
 p  in x = 1
    2

+ sum  oo  H
   -nn+1 = ln (2)
n=12
(629)

pict

By combining, we get

+ oo 
 sum  Hn-(nHn---2)= ln2(2)
n=1     n2n
(633)

Calculation of fk(1)
 p 2 for k+ p = 3

pict
We can then get
   + sum  oo    (      7)
96    Hn--nH2nn--8- = p2ln(2)
 7 n=1    n 2
(636)

Calculation of fkp (1)
   2 for k+ p = 4  To sum up :

pict
pict

We can combine those results and obtain

+ sum  oo  H2 (nH - 5)    17p4   35
   --n--2nn----= - ----+  --ln(2)z(3)
n=1     n 2         288    8
(645)

13.3.3 With x = -1

The convergence for the serie is no problems for p > 2,  we then get

+ sum  oo  (-1)nH     5
   ----2--n = -- z(3)
n=1   n        8
(646)

+o o     n+1
 sum   (-1)---Hn-=  1z(3)
n=1  (n + 1)2     8
(647)

For p = 1  we find that

pict

Which gives us

  2  + oo     n+1
p--=  sum   (--1)---(2n-+-1)Hn-
12   n=1     n (n+ 1)
(650)

Those equalities are justified because alternating series converges and with of theorem like Tauber, we conclude.
The two following equalities are formally obtained with Maple, they are satisfied numerically but I can  not yet justify them.

pict

which gives by subtracting them and with nHn - 1 = nHn- 1

pict

We also have

+o o     n
 sum   (--1)-H3n= - -p4 + 9z(3)ln(2)- p2ln2(2)+ ln4(2)
n=1    n        144   8           8           4
(654)

13.3.4 With x = i

We then get by considering the real and imaginary parts

pict
pict

The convergence is assured by making f11 (ix)  for x < 1  real and by passing through the limit in x = 1  with Abel's lemma.

With k = 1,p = 2  We get taking into account the value of L3(i)  and Landen's equality in x = i,

pict

But, we have proved in [12] (calculation of I(2)
 4  )

  (  (     ))
       1--i      35z(3)  5p2ln2-  ln32
 R  L3    2     =   64  -   192  +  48
(658)

which gives us

+ sum  oo  (- 1)n H2n  23z(3)
   -----2--- = ------- pG
n=1    n         16
(659)

and allows us to find with  sum +o o  (- 1)n   3
  n=1 -n3--= -4z (3)

           (          )
+ sum  oo  (- 1)n+1 2nH2n + 23
   ------------------6--= 2pG
n=1         n3
(660)

With k+ p = 4  The calculations are a bit more complicated, we use this time the equality (658), then the inverse formula for the polylogarithm of order 4  which gives

              (     )      2  2         4    4     (  3         3  )
L4(1 -i) = - L4 1+-i  + 11p-ln-2 + 1313p-- ln-2 - i 7p--ln2-+ p-ln-2-
                 2         768     92160    384       256      64
(661)

as well as

            4                           k
L4(i) = --7p---+ ib(4) o`u b (n) =  sum --(--1)-n
         11520                  k>0(2k+ 1)
(662)

  (   (1-+-i))    ln42   5p2ln2-2  -5   (1-)   343p4
 R  L4    2     =  96 -    768  + 16L4   2  + 92160
(663)

which correspond to the calculation of I(3)
 4  in my paper ”formules BBP”, we get

pict

We apply the same substitution for k = 2,p = 2  and k = 3,p = 1  . We then obtain by considering the real part (the imaginary part does not give anything useful)

pict
pict

By combining the different equations obtained, we can deduce

+ sum  oo      (8H3    6H2    3H  )    127p4   93z(3)ln 2   p2ln2 2  ln42
   (-1)n  ---n- --2n + --n3-  = ------ + ----------  ------+ ---- - 4G2
n=1        n     n      n        1440       8         2       4
(667)

13.3.5 With          V~ -
x = 1+ i--3
    2    2

We use in this case the following result.
If we let

pict

(mgl means "multiple by Glaishers” and mcl means ”multiple by Clausen”) then

pict

where Bn  is the nth polynomial by Bernoulli.

Then the duplication formula

                   1    (  )
Ln (z)+ Ln(- z) =-n-1Ln  z2
                 2
(672)

with      ip
z = e 3   which allows to express    ( 2ip)
Ln  e 3 and    ( 4ip )
Ln  e 3 with the help of mgl (n)  and mcl(n)  .

Note 25 the calculation of   (   )
L3 e ip3- = z(3) + 5ip3
           3    162   gives the serie

+ oo    (np)     3
 sum  sin--3- = 5p-
n=1   n3     162
(673)

which can also be written

                                       V~ 
1   1-  -1   1-  -1   1-  -1-        5-3- 3
1 + 23 - 43 - 53 + 73 + 83- 103 - ...= 243 p
(674)

We then obtain the following results :

With k+ p = 2

 + sum  oo  Hn   (np-)    p2-
     n cos  3   = -36
n=1
(675)

the convergence of this serie is justified by summation in different parts.
We also obtain (under the reservation of convergence, but it is at best very slow that it is hard to check ! ! !)

+ sum  oo  Hn   (np )
   -n-sin  3-- = -mcl (2)
n=1
(676)

but

          + sum  oo    (np)
mcl(2) = -   sin--3-
          n=1   n2
(677)

where

pict

or

   H     H    H    H    H     H
0 =--1 - -3-- --4+ --6+ --7-  -8-- ...
    2    4     5    6    7    8
(679)

With k+ p = 3

+o o       (  )
 sum   H2n-sin  np- = p3-
n=1 n       3    36
(680)

which gives us

                                           V~ 
H21-  H22-  H24   H25-  H27  H28   H29-     p3-3-
1  +  2 -  4  - 5  +  7 +  8  -  9 ...=  54
(681)

the convergence of this serie is justified by the summation by parts.

+ oo       (   )
 sum  Hn-sin  np- = 11-p3
n=1 n2     3     324
(682)

which gives

                                  V~ -
H1-   H2-  H4-  H5-  H7-       11--3 3
 1 +  22 - 42 - 52 +  72 + ...=  486 p
(683)

we also have

1                + sum o o  Hn   (np )
3 pmcl(2)+ z(3) =    n2-cos -3-
                 n=1
(684)

With k+ p = 4  we obtain the following formula

+ sum o o  Hn   (np )   17p4
    n3-cos -3- =  4860-
n=1
(685)

the other make some mcl(2),  mcl (4)....  intervine
For example

pict

which gives with

              (  )
        + sum  oo  sin-np3
mcl(4) =      n4
        n=1
(688)

pict

13.3.6 With           V~ 
x = -12 + i23-

By using the equality

   (      V~ -)       (       V~ -)
     3  i--3           1   i-3-   p2-  ipln(3)
L2   2-  2    = -L2  - 2 +  2   + 18 -    3
(690)

and it's conjugate, we have

pict

whose convegence is assured by packets.

By combining with (675), we obtain

  2     +o o    (    (     )      (   ))
ln-(3)=  sum   Hn- 2 cos  2np- - 5 cos  np-
  4     n=1 n          3           3
(692)

13.3.7 With x = 1+  i
    2   2

For k+ p = 2  We get with  1
f1  the two following formulae which are remarquables

pict
pict

and with  2
f0 ,  the two equality

pict
ln2(2)  p2   p ln (2)       + sum  oo  H2n  (np )
--8---- 96-+ ---8-- + G =    2n2-sin  -4-
                          n=1
(696)

With the functions  k
gp,  we have

pict

With k+ p = 3  The function g30  gives immediatly

 13       p2ln(2)  ln3(2)  + sum  oo  H3n    ((n + 1)p)
-16z (3) + --192--+ --48--=     -n+21sin  ----4---
                           n=1 2
(698)

 2
f1  and   1
f2  gives

pict

13.4 Generalisation

13.4.1 Euler's sums

If we define H(qn)= 1 + 1q +-1q + ..+ 1q
          2   3       n  (partial sum of z(q)  ), we have Euler's theorem.
If       + sum  oo   (q)
Sp,q =   H-np-,
      k=1 n  then Sp,a + Sq,p  is expressed with the help of z(p),z(q)  and z(p+ q)

We have some relations with the polylog : + sum  oo   (q)    Lq (x)
   Hn  xn =-1--z-
k=1  . By integrating we have for example

  oo 
 sum  H(2n)= 5z (3)
n=1 2nn  8
(701)

13.4.2 A formula combining Harmonic and combination

Here is a little serie that I've found recently in novembre 2001, mixing the combinations and the sums harmonic ! We can maybe find a more simple proof, but I do find this one quite elegant in the end.

Ok, it's a particular case, I don't know if we can find other series of this kind (and numerically I have yet to find one), but it is maybe worth the effort of searching!

Proposition 26 With Cn2n  and the harmonic sums :
If we let H(n) =  sum nk=1 1k  , we get

        (           )
 sum  oo  H(n) 18   --9---    2
    Cn2n   n -  2n+ 1  = p
n=0
(702)

Proof. Let an = HC(n2nn)n  . Then

pict

hence

 sum  oo    n   sum  oo      n+1   sum  oo -H(n)xn+1-    sum  oo ------xn+1------
    anx =     an+1x    =    2Cn2n(2n + 1) +   2Cn2n(n + 1)(2n + 1)
n=1       n=0           n=0              n=0
(704)

hence in particulae in x = 1  and by regrouping the two series containing some H(n)  , we obtain

 oo  sum  H(n) (1       1    )    oo  sum          1
   -Cn--  n-- 2(2n-+-1)  =    2Cn-(n+-1)(2n+-1)
n=0  2n                   n=0  2n
(705)

which simplify a bit more the work !

The trick     1         2      1
(n+1)(2n+1) = (2n+1)- (n+1)  unfortunatly does not seem useful because the serie in  1
n+1  is difficult to calculate it seems to me.... Let us find another way.

Note that we have                 V~       V~ 
 sum o o n=1 22nnC-n1xn =-xa V~ rcsin(-x)
        2n          1-x  i.e.  sum o o n=1 22nnC-n1x2n = xa V~ rcsin(x2)= f (x)
        2n        1-x  .

Hence

pict

using integration by parts. Hence since  integral              [V ~ ------       ]    integral 
 0xua V~ r1cs-inu(2u)du =  -  1- u2arcsin (u) x0 + 0xudu  using integration by parts, just simply,

pict

on the convergence radius of (x < 1  ).

We now need to properly integrate this serie to find a term in  1
(n+1)-   :

 sum  oo         2n                integral  x(           )                    2
    -------2---------x2n+2 =     a V~ rcsin(u)- u  du = 1arcsin2(x)-  x-
n=1 2(n + 1)(2n + 1)Cn2n         0     1- u2           2            2
(706)

let us add the term for n = 0   :

 sum  oo ------22n------- 2n+2   1     2
   2(n +1)(2n+ 1)Cn x     = 2 arcsin (x)
n=0               2n
(707)

We now use 705 to obtain

 oo  sum  H(n) (1       1    )          (1 )   p2
   --n--  --- --------  = 2arcsin2  -  = ---
n=0 C2n   n   2(2n + 1)             2    18
(708)

hence finaly

  oo      (           )
 sum  H(n)-  18-  --9--- = p2
n=0 Cn2n   n    2n + 1
(709)

 _

13.4.3 An other formula

We also find some formulae with some special harmonic sums such as Bradley's formula :

      sum  oo     n      sum n
2G =    ----2------   ---1--
     n=0(2n + 1)Cn2n k=02k + 1
(710)

or even this representation of G  which present some troubling similarity with the previous one !

     sum  oo  (--1)n+1-n sum -1-(--1)k
G =       n       2k + 1
    n=1        k=0
(711)

The proof is available in [10].

13.4.4 Harmonics of harmonics !

According to the Gradshteyn [9] (1.516), we can immediatly obtain

          oo             n        k
ln3(2) = 6  sum ----1----- sum   -1--- sum  -1
        n=1 (n+ 2)2n+2k=1 k+ 1m=1 m
(712)


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