By Maria Jose Leon-Sotelo Esteban on
Tuesday, July 31, 2001 - 05:47 am:
1/1 + 2/(2+3) + 3/(4+5+6) + 4/(7+8+9+10)+
5/(11+12+13+14+15) +... = ?
Thank you
By Arun Iyer on Tuesday, July 31, 2001 -
08:52 am:
well,
you can write the above series as,
| 2× |
lim
n®¥
|
|
n
å
r=1
|
1/(r2+1)
|
then after that i got stuck and only thing i could think of was
to replace the above limit by an integration sign as...
2×ò0¥ 1/(x2+1) dx=2×[arctan x]0¥=2×(p/2) = p
believe me i am not completely satisfied with this
result...
please do correct me..
love arun
By David Loeffler on Tuesday, July 31,
2001 - 11:20 am:
It is in fact pcothp-1.
The proof is by complex analysis and is rather devious. We actually prove that
, by considering the singular
points of the function f(z)=pcot(pz)/(z2+1).
By Arun Iyer on Tuesday, July 31, 2001 -
01:27 pm:
David,
Can we get the series of pcothp
instead of prooving it the other way round?
i believe this would be a little easier.
love arun
By Arun Iyer on Tuesday, July 31, 2001 -
01:28 pm:
David,
can you tell me what a singular point in a function means?
love arun
By Arun Iyer on Tuesday, July 31, 2001 -
01:34 pm:
Does your proof of
deal with complex
integration?
i have done some work on it by studying from the sites and now i
have a fair bit of knowledge on it.Though compared to what you
discuss it seems quite little.
well when you say singular points do you mean poles of the
function??
love arun
By Brad Rodgers on Tuesday, July 31, 2001
- 06:36 pm:
A way that might be easier (I haven't been able to finish it)
is to break the sum up using partial fractions as
| 2 |
¥
å
1
|
1/(r2+1)=i |
¥
å
1
|
1/(r+i)-1/(r-i)
|
It is also an elementary result of the Psi function that
|
n
å
1
|
1/(r+b)=y(n+b+1)-y(b+1)
|
Using this to evaluate the first sum, we get that
| 2 |
¥
å
1
|
1/(r2+1)=i(y(1-i)-y(1+i))
|
There was a discussion recently (entitled series for Psi) that
talked of a relation between Psi and coth I think, but I think
this page is being moved to Asked Nrich, or is at least I can't
seem to find it. This could provide an easier way to prove the
result though.
Brad
Series for
Psi
By Brad Rodgers on Tuesday, July 31, 2001
- 08:20 pm:
The formula I was looking for was
y(-z)=y(z)+p×cot(pz)+1/z
=y(z+1)+p×cot(pz)
As y(z+1)=1/z+y(z). Also, using that result, and adding -1/z to both
sides
y(1-z)=y(1+z)+p×cot(pz)-1/z
Subbing in z=i,
y(1-i)=y(1+i)+p×cot(ip)+i
Therefore
i(y(1-i)-y(1+i))=ip×cot(ip)-1
As coth(x)=i×cot(i x); we are given the result
p×coth(p)-1
without using comlex analysis.
Does anyone know how to prove y(-z)=y(z)+p×cot(pz)+1/z though. < BR/> I remember that the use of series was used in the first discussion, but the < BR/> series initially used wasn't ever proven.
By David Loeffler on Wednesday, August
01, 2001 - 11:39 am:
Possibly one starts from G(x)G(-x)=psin(px) or whatever it is?
By David Loeffler on Wednesday, August 01,
2001 - 01:37 pm:
Arun,
A singular point is a point where a function fails to be
analytic. A pole is a type of singular point (not all singular
points are poles). However all the singular points in my method
are actually poles of order 1.
The function I mentioned has singular points at every real
integer, and it can be proved that at n, its residue is
1/(n2 +1).
However, it also has two singular points at ±i.
Now, we can show that if we take a very large square in the
complex plane, centred at the origin and having corners
(±(n+1/2), ±(n+1/2)), then as n-> infinity, the
integral around this square as a contour tends to zero. Hence we
see that the sum of all the residues at the integers + the sum of
the residues at ±i = 0. So we just calculate those two
residues and the series drops out.
We can use this technique on a lot of similar series. My complex
analysis textbook has pages and pages of them.