Need a proof about convergence of a series

[AdrianZ]

I was reading a book about Calculus that I came to a problem that the author claimed convergence of a series won't change if we subtract a finite number of its terms from it, It seems to be intuitively clear, but I need a proof. so please Prove that the convergence/divergence status of a series won't change if we add/subtract a finite number of terms to/from it. please make it as rigorous as possible using only single-variable Calculus theorems.

Thanks in advance.

 

[CompuChip]

You can easily show it from the definition:
\forall \epsilon > 0, \exists N > 0: \ldots

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).

 

[AdrianZ]

Yea, actually I was thinking about the same way of proving this, I said when we say a series converges it means \forall\epsilon>0\existsN>0 s.t n>N \Rightarrow |S(n) - L|< \epsilon 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.

 

[CompuChip]

OK, let's define
S_n = \sum_{k = 1}^n a_k
and assume that there exists some real number L
\lim_{n \to \infty} S_n = L
(this is actually what it means to say that \sum_{k = 1}^\infty a_k has a limit).

Your question is: does there exist a number L' such that
\sum_{k = \ell + 1}^\infty a_k = L&#039;
that is, such that
\lim_{n \to \infty} \left( S_n - S_\ell \right) = L&#039;

Actually, since the second term is just a constant with respect to n, you can use that
\lim_{n \to \infty} (a_n + c) = \left( \lim_{n \to \infty} a_n \right) + c
and infer that L&#039; = L - S_\ell.

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

 

[AdrianZ]

Yea, actually I see the idea, but that only proves the theorem for a specific case, I mean in that case you've dropped all the terms before l+1 and have proved that the series is still convergent, am I right? can we do that in any random way we like? I mean can we drop any terms we like arbitrarily and still come to a new convergent series which its limit is the same? (provided we do that for a finite number of times)

 

[HallsofIvy]

Yea, actually I see the idea, but that only proves the theorem for a specific case, I mean in that case you've dropped all the terms before l+1 and have proved that the series is still convergent, am I right? can we do that in any random way we like? I mean can we drop any terms we like arbitrarily and still come to a new convergent series which its limit is the same? (provided we do that for a finite number of times)

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 finite and removing it would change the sum by finite amount.

 

[AdrianZ]

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 finite and 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?

 

[icystrike]

\mid S_{l} - L&#039; \mid &lt; \frac{\epsilon}{2} \forall l&gt;N

\mid S_{n} - S_{l}\mid\ =\mid S_{n} -L&#039; + L&#039; - S_{l}\mid\ \leq \mid S_{n} -L&#039; \mid + \mid L&#039; - S_{l} \mid &lt; \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon
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).
\mid S_{n} - S_{l}\mid\ =\mid S_{n} -L&#039; + L&#039; - S_{l}\mid\ \leq \mid S_{n} -L&#039; \mid + \mid L&#039; - S_{l} \mid &lt; \mid S_{n} - L&#039; \mid + \frac{\epsilon}{2}
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') ...