Please help me to evaluate the series,

Must go to my lecture!

Sorry to take so long to get back.

David, the fact that the sum can be expressed like that is indeed not a coincident; it deals with the fact that the sum is equal to

which, by the reflection formula for psi (e.g. y(x)-y(1-x)=pcot(px)) is equal to

p/8 cos(7p/8).

If anyone wants more details, I can provide them, but for now I won't explain much more because I have found a better way.

For the sine product,

sin(px)=x

¥ Õ n=1

(1-x2/n2)

,
if we take logarithms and differentiate, we get

pcot(px)=1/x-

¥ å n=1

2x/(n2-x2)=1/x+

¥ å n=1

1 n+x

-

1 n-x

Therefore,

p/8cot(p/8)=1+

¥ å k=1

1 8k+1

-

¥ å n=1

1 8n-1

=

¥ å n=1

1/(8n-7)-1/(8n-1)

As required. To prove that cot(p/8)=1+Ö2, consider a 1, 1, Ö2 right triangle, and use the angle bisector theorem twice.
As an afterthought, this method, though probably shorter on paper is really a roundabout method. To prove the sine expansion, which is the basis for this proof, one has to prove the reflection formula for gamma, and generally, the result that about harmonic sums and the psi function is used as well.

Brad

What is the 'reflection' of a function?

Yatir

Brad,
the solution by sine product was a wonderful technique....

also,i am not much aware of the psi function and its properties...please could you explain it a bit...!!!

love arun

David,
i would like to know how you expressed the series as integral .......

love arun

Yatir, reflection formulas are things used to find the value of a function when it goes past a given point. Sometimes, they are called analytic continuations. For the gamma function, the reflection formula is

• y(1-x)-y(x)=pcot(px)
• y(1+x)=1/x+y(x)

Recall that 1/n=ò₀¹ x^n-1 dx. Now your series is

¥ å n=0

[(8n+1)-1-(8n+7)-1]

=

¥ å n=0

ò01(x8n-x8n+6)dx

=ò01

¥ å n=0

(x8n-x8n+6)dx

(this step is justified by monotone convergence theorem)

=ò₀¹(1-x⁶)/(1-x⁸)dx

Kerwin

Here are the proofs for a) and b). Don't look if you want to prove them on your own:

a) We know

(can you see why d/dx(ln(G(1-x)))=-y(1-x)?). Therefore,

y(1-x)-y(x)=pcot(px)

b) This one is easier: taking logs on the relation G(x+1)=xG(x) and differentiating, we get the result.
Now, forget everything I've typed above on the Psi function for this series; most of it was wrong (I typed it at school and thought I cold remember my work from home). Can you see how we derive

y(N+1+x)=y(x)+

N å n=0

1/(n+x)

Therefore

y(N+1/8)/8=y(1/8)/8+

N-1 å n=0

1/(8n+1)=y(1/8)/8+

N å r=1

1/(8r-7)

(1)
Similarly,

y(N+7/8)/8=y(7/8)/8+

N å r=1

1/(8r-1)

(2)
Therefore subtracting (2) from (1),

[y(N+1/8)-y(N+7/8)]/8=

1 8

(y(1/8)-y(7/8))+

N å r=1

1/(8r-7)-1/(8r-1)

... (*)
Now, we want to prove that

lim N®¥

y(N+1/8)-y(N+7/8)=0

.
A slightly non-rigorous proof is as follows (it is non-rigorous only because I list a couple intuitional statements without proof):

We know

i) y (x) is a monotonically increasing function for x > 0(look at a graph, or consider that the function is a relatively simple curve, with no twists or turns, and recall the series value for it)

ii) because it's monotonically increasing, what we are trying to evaluate will be less than zero

iii) Once again because it's monotonically increasing, y(N+1/8) > y(N-1+7/8).
So we know

0 > y(N+1/8)-y(N+7/8) > y(N-1+7/8)-y(N+7/8)=-1/(N-1+7/8)

Clearly, as N®¥, y(N+1/8)-y(N+7/8) must be bounded arbitrarily close to zero, and therefore, our limit tends to zero.
Then, putting this back into (*), we get

1 8

(y(7/8)-y(1/8))=

N å r=1

1/(8r-7)-1/(8r-1)

Which, by the reflection formula, is equal to

p 8

cot(p/8)

That was a pretty long post, and I'm sure I missed out on some details; post back if anything was hard to understand.

As a sidenote, the function I have referred to as the psi function here is sometimes called the digamma function.

Brad

By the way, the nice thing about using the psi function this way is that the technique generalizes nicely for other series composed of a number of harmonic series.

Kerwin,
i didn't get last two steps of your proof...and what is monotone convergence theorem???

love arun

Monotone convergence theorem states that if each f_n is measurable with 0 £ f_n £ f_n+1, then

lim

òfn(x)dm®

lim

f(x) dm

Do you know anything about measure (e.g. Lebesgue measure/integral)? If you don't know anything about measure, don't worry. Measure can be thought of as a generalisation of the Riemann integral, which behaves much nicer in limiting process. The theorem says (in terms of Riemann integration) that if 0 £ f_n £ f_n+1 and each f_n is integrable, then the integral of the limit of f_n is the limit of the integral of f_n.

The last step is just summing a GP (0 £ x < 1), and what happens at the end-point x=1 does not matter (since a point won't contribute to the integral).

Sorry!!to answer so late i am quite busy with my exam studies....

Brad,
thanks for pointing out that silly error i had done and also thanks for the explanation....

Kerwin,
i don't know anything about measures or riemann integration....i will look through the archive you have suggested but it may take me sometime to understand.....

Thanks everyone for all the explanation!!!
love arun