Question

How is it?
\[\sum_{n\geq 0}(\dfrac{1}{n+1})^2 =\sum_{n\geq 1}\dfrac{1}{n^2}\]
Please, help

Answer 1

@reemii

Answer 2

If you write the first terms of both series, you will see something...

Answer 3

Other method (more formal than just "seeing"): In the LHS, set \(k=n+1\) and see how the expression becomes in terms of \(k\).

Answer 4

Yeah.... I "see" it now. Thank you so much. One more question

Answer 5

why \[\int_0^1\int_0^1 \dfrac{1}{1-xy}dxdy\] is a square?

Answer 6

The integral is not a square. Computing the integral gives a value.

It is the "domain of integration" that is a square. The domain of integration is described at the limits of the integral symbols \(\int\).

Answer 7

but when x=1,y=1, the function is undefined, right?

Answer 8

It is possible that the value of the integral is \(+\infty\).
Yes, the function can be undefined at the limits, but that means that you have to compute a "generalized integral". Just as in this case:
\[ \int_0^1 \frac1{\sqrt x}\, dx \]
Not defined at x=0 but the integral has a finite value.

Answer 9

I read three time pi^2/6 with its proof. I don't get why they have to use the transformation to calculate \(\zeta (2)\). Can you explain me?

Answer 10

http://link.springer.com/chapter/10.1007/978-3-662-44205-0_9#page-1

Answer 11

I can only see that they succeed in computing it..
How would you evaluate the integral?

Answer 12

I understand all of the proof, just want to make clear why they use u,v plane to calculate the integral instead of x,y plane.

Answer 13

Is it not that because the function is undefined when x=y=1. (As they states, it is a hole there)?

Answer 14

They make a change of variables -you might choose another if you want- that leads to a cool expression.. and for which the integral can be computed.

Maybe if you're lucky you can find another change of variable that works fine.

Answer 15

hehehe.. NO, I am not that good, I just want to completely understand it.

Answer 16

After transfer to uv plane, the function becomes \(\dfrac{1}{1-u^2+v^2}\)
Hence when \(u^2-v^2=1\) , it is undefined, but at that time, the function is out of the new domain integration, right?

Answer 17

No, it is not, when u =1, v =0 , we have u^2-v^2 =1, still undefined.

Answer 18

That is why I want to know the reason they have to transform to u, v plane.

Answer 19

:-o laptop crash..
Actually, you're right to be careful about the domain of the function. Here, it is an improper integral.
https://en.wikipedia.org/wiki/Improper_integral

The integral still makes sense.
The change of variable doesn't change the property of the integral being improper. It just changes the expression of the function to integrate. (the limits too change...)

Since the authors are professional mathematicians, they do not take the time to make the details about improper integrals that students would make. (computing limits)

Answer 20

Thanks for the link. I do appreciate. Let me read and digest it first. :)