Can someone verify the infinit sum for 4*Pi^2/3
Gerry
sum_m=0^Inf(4/(4*m^3-m)+(2*H(m-3/2)+4*log(2))/(m+2*m^2))
H(n)= Harmonicnumbers
I could not find any formulas using H(m-3/2) at
http://mathworld.wolfram.com/HarmonicNumber.html
Thanks
Gerry
Henry
> Hi the sum is :
>
> sum_m=0^Inf(4/(4*m^3-m)+(2*H(m-3/2)+4*log(2))/(m+2*m^2))
>
> H(n)= Harmonicnumbers
Problems:
(1) Harmonic numbers are usually only defined for positive integers
(2) When m=0, the donominators both seem to be 0
Gerry
Well yes usually they are but using a series expansion I get these
results.
Take
G(n)= sum_m=n^Inf(4/(4*m^3-m)+(2*H(m-3/2)+4*log(2))/(m+2*m^2))
I get that:
G(0)= 4*Pi^2/3
G(1)= Pi^2/3
G(2)= Pi^2/3 - 4/3
G(3)= Pi^2/3 - 28/15
G(4)= Pi^2/3 -136/63
G(5)= Pi^2/3-2216/945
...
So G(0) is not corerct?
Kaba
> On Feb 3, 10:07 am, Gerry <gerry...@gmail.com> wrote:
> > Hi the sum is :
> >
> > sum_m=0^Inf(4/(4*m^3-m)+(2*H(m-3/2)+4*log(2))/(m+2*m^2))
> >
> > H(n)= Harmonicnumbers
>
> Problems:
> (1) Harmonic numbers are usually only defined for positive integers
Harmonic numbers can be generalized to reals by the digamma function.
But maybe your "usually" covers that:)
Rob Johnson
There are many parts of sizeable length to this proof. I have tried
to break it up into more easily digestible pieces. I hope it can be
followed without too much trouble.
Extended Harmonic Numbers
-------------------------
In <http://www.whim.org/nebula/math/extendharm.html>, I show that
H(-1/2) = -2 log(2) [1]
Using [1] and
H(k) = H(k-1) + 1/k [2]
we can derive
H(m-3/2) = 2 H(2m-3) - H(m-2) - 2 log(2) [3]
Using [1] and [2], we get that
H(-3/2) = 2 - 2 log(2) [4]
Differentiating the definition of H(x) in the article cited above,
we also get that
oo
--- 1
H'(-3/2) = > ---------
--- (k-3/2)^2
k=1
oo
--- 1
= 4 + > ---------
--- (k-1/2)^2
k=1
oo
--- 1
= 4 + 4 > --------
--- (2k-1)^2
k=1
= 4 + 3 zeta(2)
= 4 + pi^2/2 [5]
Summation Summary
-----------------
I computed the desired sum in three steps.
1. summing the first two terms separately since the first term is
technically singular and the second term doesn't work well with
the formula for H(m-3/2) that I would like to use.
2. summing 4/(4m^3 - m) since this is easy to do using partial
fractions and the series for log(2).
3. summing the harmonic number terms.
1. Summing the First Two Terms
------------------------------
We are trying to compute
oo
--- 4 2 H(m-3/2) + 4 log(2)
> -------- + --------------------- [6]
--- 4m^3 - m 2m^2 + m
m=0
Technically, the first term (m = 0) is singular. However, if we
consider the first term as a continuous function of m near 0, we
can compute the limit as m -> 0.
Use [4] and [5] to expand 2 H(m-3/2) + 4 log(2) about m = 0:
4 2 H(m-3/2) + 4 log(2)
-------- + ---------------------
4m^3 - m 2m^2 + m
1 4
= ------- ( ---- + 2 (2 - 2log(2)) + 2 (4 + pi^2/2) m + O(m^2) + 4 log(2) )
m(2m+1) 2m-1
1 4
= ------- ( ---- + 4 + (8 + pi^2) m + O(m^2) )
m(2m+1) 2m-1
1 8m
= ------- ( ---- + (8 + pi^2) m + O(m^2) )
m(2m+1) 2m-1
1 8
= ---- ( ---- + (8 + pi^2) + O(m) )
2m+1 2m-1
-> pi^2 [7]
as m -> 0.
At m = 1, use [1] to get
4 2 H(m-3/2) + 4 log(2)
-------- + ---------------------
4m^3 - m 2m^2 + m
4
= - [8]
3
Thus, we can use [7], [8], and [3] to obtain
oo
--- 4 2 H(m-3/2) + 4 log(2)
> -------- + ---------------------
--- 4m^3 - m 2m^2 + m
m=0
oo
4 --- 4 4 H(2m-3) - 2 H(m-2)
= pi^2 + - + > -------- + -------------------- [9]
3 --- 4m^3 - m 2m^2 + m
m=2
2. Summing 4/(4m^3 - m)
-----------------------
By partial fractions, we get that
1 1 2 1
------------- = ---- - -- + ---- [10]
m(2m-1)(2m+1) 2m-1 2m 2m+1
Using [10] we get
oo
--- 4
> --------
--- 4m^3 - m
m=2
1 2 2 2 2 2 2
= 4 ( - - - + - - - + - - - + - ... )
3 4 5 6 7 8 9
2 2 1 2 2 2 2 2 2 2 2 2
= 4 (-( - - - + - ) + - - - + - - - + - - - + - - - + - ... )
1 2 3 1 2 3 4 5 6 7 8 9
4
= 4 (- - + 2 log(2) ) [11]
3
Using [9] and [11], we get
oo
--- 4 2 H(m-3/2) + 4 log(2)
> -------- + ---------------------
--- 4m^3 - m 2m^2 + m
m=0
oo
--- 4 H(2m-3) - 2 H(m-2)
= pi^2 - 4 + 8 log(2) + > -------------------- [12]
--- 2m^2 + m
m=2
3. Summing the Harmonic Number Terms
------------------------------------
Note that
H(2m-3) - H(m-2)/2
m-2
--- 1
= > ---- [13]
--- 2k+1
k=0
So that
oo
--- 4 H(2m-3) - 2 H(m-2)
> --------------------
--- 2m^2 + m
m=2
oo m-2
--- --- 1
= 4 > > -------------
--- --- (2k+1)(2m+1)m
m=2 k=0
oo oo
--- --- 1
= 4 > > -------------
--- --- (2k+1)(2m+1)m
k=0 m=k+2
oo oo
--- 1 --- 1 1
= 8 > ---- > -- - ----
--- 2k+1 --- 2m 2m+1
k=0 m=k+2
oo
--- 1 |\1 x^{2k+3}
= 8 > ---- | -------- dx
--- 2k+1 \| 0 1+x
k=0
|\1 1+x x^2
= 4 | log( --- ) --- dx
\| 0 1-x 1+x
= 4 - 8 log(2) + pi^2/3 [14]
Things are getting long-winded enough; I defer the justification of
the last step of [14].
Final Result
------------
Combining [12] and [14], we get
oo
--- 4 2 H(m-3/2) + 4 log(2)
> -------- + ---------------------
--- 4m^3 - m 2m^2 + m
m=0
= 4/3 pi^2 [15]
As foretold.
The Last Step of [14]
---------------------
Before we get to the main part of the derivation, we will need a
couple of integrals, namely [17] and [20] below.
First note that integration by parts yields
|\1 k
| x log(x) dx
\| 0
x^{k+1} |1 |\1 x^{k+1} 1
= ------- log(x) | - | ------- - dx
k+1 |0 \| 0 k+1 x
1
= - ------- [16]
(k+1)^2
Thus,
|\1 log(x)
| ------ dx
\| 0 1-x
|\1
= | log(x) ( 1 + x + x^2 + x^3 + x^4 + ... ) dx
\| 0
1 1 1 1 1
= - ( --- + --- + --- + --- + --- + ... )
1^2 2^2 3^2 4^2 5^2
pi^2
= - ---- [17]
6
Furthermore, we will need the following
|\1 log(x)
| ------ dx
\| d 2-x
|1 |\1 log(2-x)
= -log(x) log(2-x) | + | -------- dx
|d \| d x
|\2-d log(x)
= log(2) log(d) + | ------ dx [18]
\| 1 2-x
Adding the integral on the left of [18] to both sides, and then
substituting x -> 2x, we get
|\1 log(x)
2 | ------ dx
\| d 2-x
|\2-d log(x)
= log(2) log(d) + | ------ dx
\| d 2-x
|\1-d/2 log(x) + log(2)
= log(2) log(d) + | --------------- dx
\| d/2 1-x
= log(2)^2 + log(2) log(1-d/2)
|\1-d/2 log(x)
+ | ------ dx [19]
\| d/2 1-x
Sending d -> 0 and using [17] and [19], we get
|\1 log(x)
| ------ dx
\| 0 2-x
1 2 pi^2
= - log(2) - ---- [20]
2 12
Now, on to the last step of [14]
|\1 1+x x^2
| log( --- ) --- dx
\| 0 1-x 1+x
|\1 1
= | log(1+x) ( --- + x - 1 ) dx
\| 0 1+x
|\1 1
- | log(1-x) ( --- + x - 1 ) dx
\| 0 1+x
|\1 |\1 1 2
= | log(1+x) d log(1+x) + | log(1+x) d( - x - x )
\| 0 \| 0 2
|\1 |\1 1 2
- | log(1-x) d log(1+x) - | log(1-x) d( - x - x )
\| 0 \| 0 2
1 2 |1 1 2 |1 |\1 1 2 1
= - log(1+x) | + ( - x - x ) log(1+x) | - | ( - x - x ) --- dx
2 |0 2 |0 \| 0 2 1+x
|\1 log(x) 1 2 |1-d |\1-d 1 2 1
- | ------ dx - ( - x - x ) log(1-x) | - | ( - x - x ) --- dx
\| 0 2-x 2 |0 \| 0 2 1-x
1 2 1 |\1 1 3
= - log(2) - - log(2) - | - ( x - 3 + --- ) dx
2 2 \| 0 2 1+x
pi^2 1 2 1 |\1-d 1 1
+ ---- - - log(2) + - log(d) - | - ( 1 - x - --- ) dx
12 2 2 \| 0 2 1-x
1 2 5
= - log(2) - 2 log(2) + -
2 4
pi^2 1 2 1 2 1
+ ---- - - log(2) + - d - -
12 2 4 4
pi^2
= 1 - 2 log(2) + ---- [21]
12
Rob Johnson <r...@trash.whim.org>
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Rob Johnson
This is not a duplicate post, I reworked the last step of [14] to be
a simpler and not use a limiting argument (that I glossed over in my
previous post).
Using [1] and
we can derive
-> pi^2 [7]
as m -> 0.
Using [10] we get
So that
= 4/3 pi^2 [15]
As foretold.
Thus,
|\1 |\1 1 2 1
= | log(1+x) d log(1+x) + | log(1+x) d( - x - x + - )
\| 0 \| 0 2 2
|\1 |\1 1 2 1
- | log(1-x) d log(1+x) - | log(1-x) d( - x - x + - )
\| 0 \| 0 2 2
1 2 |1 1 2 1 |1 |\1 1 2 1 1
= - log(1+x) | + ( - x - x + - ) log(1+x) | - | ( - x - x + - ) --- dx
2 |0 2 2 |0 \| 0 2 2 1+x
|\1 log(x) 1 2 1 |1 |\1 1 2 1 1
- | ------ dx - ( - x - x + - ) log(1-x) | - | ( - x - x + - ) --- dx
\| 0 2-x 2 2 |0 \| 0 2 2 1-x
1 2 |\1 1 4
= - log(2) - | - ( x - 3 + --- ) dx
2 \| 0 2 1+x
pi^2 1 2 |\1 1
+ ---- - - log(2) - | - ( 1 - x ) dx
12 2 \| 0 2
1 2 5
= - log(2) - 2 log(2) + -
2 4
pi^2 1 2 1
+ ---- - - log(2) - -
12 2 4
pi^2
Gerry
> In article <fc9db982-e8d6-49cf-bc4c-529fd9937...@y3g2000vbh.googlegroups.com>,
Hi Rob,
thank you for this wonderful proof.
Rob Johnson
|\2-d log(x)
= log(2-d) log(d) + | ------ dx [18]
\| 1 2-x
>> Adding the integral on the left of [18] to both sides, and then
>> substituting x -> 2x, we get
|\1 log(x)
2 | ------ dx
\| d 2-x
|\2-d log(x)
= log(2-d) log(d) + | ------ dx
\| d 2-x
|\1-d/2 log(x) + log(2)
= log(2-d) log(d) + | --------------- dx
\| d/2 1-x
= log(2) log(2-d) + log(d) log(1-d/2)
Note that I have corrected [18] and [19] above. I was really hoping
that someone would come up withs something a bit simpler.
Axel Vogt
I tried to cross check numerical using Maple, which writes
H(n) = Psi(n+1) + gamma, gamma = Euler's constant ~ 0.57
(and Maple does not give a numerical approximation, but it
covers parts of Rob Johnson's answer).
For m=0 the summand is Pi^2 and now summing for 1 <= m over
4/(4*m^3-m) +
(2*gamma+4*ln(2))/(m+2*m^2) +
1/(m+2*m^2)*Psi(m-1/2)
This then gives
Pi^2 + # ~ 9.87
-4+8*ln(2) + # ~ 1.54
4*gamma-4*gamma*ln(2)+8*ln(2)-8*ln(2)^2 + # ~ 2.41
S3 # unknown
where S3 =Sum(1/(m+2*m^2)*Psi(m-1/2),m = 1 .. infinity);
Now Pi^2 + 1st + 2nd + S3 ~ 13.82 + S3, however 4/3*Pi^2 ~ 13.16,
but S3 is positive as a sum of positive Reals.
What is wrong in the above?
Axel Vogt
The error is mine: -2/3*gamma-4/3*ln(2) is the first summand
in S3 (for m=1) and not very positive ...