# Sum question.

1. Feb 27, 2006

### MathematicalPhysicist

sum question.........

i have b_k=a_(n_(k)+1)+...+a_(n_(k+1))

c_k=b_1+...+b_k
let us suppose that for every term in b_k has the same sign and that the sum b_k (from n=1 to infinity) converges.
S_n=a_1+...+a_n
and n belongs to {n_k+1,...,n_(k+1)}

then i need to show that
|S_n-c_n|=|a_(n_(k)+1)+...+a_n|

i tried just opened it, and it looked disastrous ( if that's even a word in the anglo-american lexicon).

2. Feb 27, 2006

### benorin

Do you mean $$b_k=a_{n_{k}+1}+\cdots +a_{n_{k+1}}$$ ?

And, if so, I don't "get it".

3. Feb 27, 2006

### VietDao29

It's like your 700-th post here, and I guess it does not hurt much if you try to learn how to LaTeX properly, right?
I don't really get what you mean, either...

4. Feb 28, 2006

### MathematicalPhysicist

and to the other user, now it's 701, but who counts it anyway? (-:

5. Feb 28, 2006

### qtp

funny thing is you are already using latex lol just put the [ tex] [ /tex] on either end and change ( ) for { } and it's pretty much the same thing

6. Mar 1, 2006

7. Mar 2, 2006

### VietDao29

Well, I don't think I understand the problem fully, but you can try to open everything:
$$b_k = a_{n_k + 1} + ... + a_{n_{k + 1}}$$
So:
$$c_k = \sum_{i = 1} ^ k b_i = a_{n_1 + 1} + a_{n_1 + 2} + ... + a_{n_2} + a_{n_2 + 1} + ... + a_{n_3} + ... + a_{n_{k + 1}}$$
$$S_n = \sum_{i = 1} ^ n a_i$$
So that means:
$$S_n - c_n = ...$$
EDIT: But are you sure you posted this problem correctly? I don't think it's correct...

Last edited: Mar 2, 2006