- #1

- 319

- 0

Thanks in advance.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter AdrianZ
- Start date

- #1

- 319

- 0

Thanks in advance.

- #2

CompuChip

Science Advisor

Homework Helper

- 4,302

- 47

[tex]\forall \epsilon > 0, \exists N > 0: \ldots[/tex]

If you subtract M terms, basically all you have to do is shift N to N - M (or you can show that the same N will still work).

- #3

- 319

- 0

Yea, actually I was thinking about the same way of proving this, I said when we say a series converges it means [tex]\forall[/tex][tex]\epsilon[/tex]>0[tex]\exists[/tex]N>0 s.t n>N [tex]\Rightarrow[/tex] |S(n) - L|< [tex]\epsilon[/tex] where S(n) is supposed to mean the sum of any n terms of the series, then I tried to use the triangle inequality and show that if S'(n) is the sum of some finite n terms of the series, then by using some algebra, we can show that they still converge for some N by definition.

let me brief it this way, we know that if n is a finite number then the Sum of any n numbers will certainly converge to a specific real number because we have no problem with summing a finite number of terms, from another theorem we know that if a series converges then we're allowed to write it in linear combinations of different convergent series. so here we have a series that is convergent based on the hypothesis and we have another series which is convergent because It's a finite sum, so any linear combinations of these two must be convergent. am I right? I see the idea, but I think It's not convincingly rigorous.

let me brief it this way, we know that if n is a finite number then the Sum of any n numbers will certainly converge to a specific real number because we have no problem with summing a finite number of terms, from another theorem we know that if a series converges then we're allowed to write it in linear combinations of different convergent series. so here we have a series that is convergent based on the hypothesis and we have another series which is convergent because It's a finite sum, so any linear combinations of these two must be convergent. am I right? I see the idea, but I think It's not convincingly rigorous.

Last edited:

- #4

CompuChip

Science Advisor

Homework Helper

- 4,302

- 47

[tex]S_n = \sum_{k = 1}^n a_k[/tex]

and assume that there exists some real number L

[tex]\lim_{n \to \infty} S_n = L[/tex]

(this is actually what it means to say that [itex]\sum_{k = 1}^\infty a_k[/itex] has a limit).

Your question is: does there exist a number L' such that

[tex]\sum_{k = \ell + 1}^\infty a_k = L'[/tex]

that is, such that

[tex]\lim_{n \to \infty} \left( S_n - S_\ell \right) = L'[/tex]

Actually, since the second term is just a constant with respect to n, you can use that

[tex]\lim_{n \to \infty} (a_n + c) = \left( \lim_{n \to \infty} a_n \right) + c[/tex]

and infer that [itex]L' = L - S_\ell[/itex].

But if you want, you can write out the limits and do it all properly.

- #5

- 319

- 0

- #6

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 963

So you are asking about cases like when we drop the first 5 terms, then terms 100- 150, then 1050 to 2000, etc. and not just "the first n terms"? That's more complicated but can still be done by dividing the terms of the series into sets- the first terms that are removed from the series, the next set of those that are removed, the third set of terms that are removed, etc. The crucial point is that there are a finite number of such sets the sum of each such set of tems is

- #7

- 319

- 0

So you are asking about cases like when we drop the first 5 terms, then terms 100- 150, then 1050 to 2000, etc. and not just "the first n terms"? That's more complicated but can still be done by dividing the terms of the series into sets- the first terms that are removed from the series, the next set of those that are removed, the third set of terms that are removed, etc. The crucial point is that there are a finite number of such sets the sum of each such set of tems isfiniteand removing it would change the sum by finite amount.

yea, that's the case I'm trying to prove it, can you give a proof of that?

- #8

- 446

- 1

[tex]\mid S_{l} - L' \mid < \frac{\epsilon}{2} \forall l>N[/tex]

[tex]\mid S_{n} - S_{l}\mid\ =\mid S_{n} -L' + L' - S_{l}\mid\ \leq \mid S_{n} -L' \mid + \mid L' - S_{l} \mid < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon [/tex]

for n and l taken to sufficiently large...(>N)

Therefore Sn-Sl is cauchy.

Now consider l is relatively small number(not taking to very large).

[tex]\mid S_{n} - S_{l}\mid\ =\mid S_{n} -L' + L' - S_{l}\mid\ \leq \mid S_{n} -L' \mid + \mid L' - S_{l} \mid < \mid S_{n} - L' \mid + \frac{\epsilon}{2} [/tex]

Therefore Sn-Sl is bounded even if l is relatively small.

Since it is bounded, and cauchy ... Therefore Sn-Sl is convergent even when l is small (finte).

Hence, as n-> infinity, Sn-Sl will go to its supremum (Sn-L') ....

[tex]\mid S_{n} - S_{l}\mid\ =\mid S_{n} -L' + L' - S_{l}\mid\ \leq \mid S_{n} -L' \mid + \mid L' - S_{l} \mid < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon [/tex]

for n and l taken to sufficiently large...(>N)

Therefore Sn-Sl is cauchy.

Now consider l is relatively small number(not taking to very large).

[tex]\mid S_{n} - S_{l}\mid\ =\mid S_{n} -L' + L' - S_{l}\mid\ \leq \mid S_{n} -L' \mid + \mid L' - S_{l} \mid < \mid S_{n} - L' \mid + \frac{\epsilon}{2} [/tex]

Therefore Sn-Sl is bounded even if l is relatively small.

Since it is bounded, and cauchy ... Therefore Sn-Sl is convergent even when l is small (finte).

Hence, as n-> infinity, Sn-Sl will go to its supremum (Sn-L') ....

Last edited:

Share: