Summation of a sequence by parts.

[r.a.c.]

I hope can someone clarify this for me.

I have a sequence f(of n) which is like this:

fn(x) = $$0-- if--x<\frac{1}{n+1}$$
is = $$sin^2(x/pi)--if--\frac{1}{n+1}<=x<=\frac{ 1}{n}$$
is = $$0--if--\frac{1}{n}<x$$

(the - are for spaces because I don't know how to do it. Nothing is negative)

Then there is a summation (sigma) of f(of n). I don't understand how can there be such a summation for sequence by parts or how it would look like. Can anyone explain please?

 

[HallsofIvy]

Given any positive x, there exist N such that x< 1/(N+1) and M such that x> 1/M. That is, the sum is a finite sum, between N and M, for every x.

For example, if x= 1/3, then, Since 1/3< 1/(0+1), f₀(1/3)= 0. Since 1/3< 1/(1+1), f<sub>1^{(1/3)= 0. Since 1/(2+1)= 1/3< 1/2, f2(1/3)= $sin^2(1/(3\pi)$. Since 1/(3+1)< 1/3= 1/3, f3}(1/3)= $sin^2(1/(3\pi))$. Since 1/3< 1/4, f4(1/3)= 0 and fn(1/3)= 0 for all larger n.

The sum is $f(x)= 2sin^2(1/(3\pi))$.

In fact, it looks to me like, if x= 1/n for some integer n, $f(x)= 2sin^2(1/(\pi x))$ while if x is any other number, $f(x)= sin^2(1/(\pi x))$ since if x= 1/n, we have both x= 1/n and x= 1/((n-1)+1) and so sum two terms, while if x is not, we have only 1/(n+1)< x< 1/n for a single n.</sub>

 

[r.a.c.]

Hmmm, I think I get it. Thanks.