Can the Infinite Series Theorem Extend to Negative Indices?

[mathsciguy]

For the series such that: \Sigma _{n=1} ^{\infty} a_n =\Sigma _{n=1} ^{\infty} b_n A certain theorem says that these series are equal even if a_n = b_n only for n>m. That is, even if two infinite series differ for a finite number of terms, it will still converge for the same sum. I am thinking that with this, we can extend the value of the starting index to include even the negative integers, and this will also extend the familiar operation that we do with finite summations where let's say:

\Sigma _{k=1} ^{5} a_k =\Sigma _{k=2} ^6 a_{k-1} To its infinite series analog:
\Sigma _{k=1} ^{\infty} a_k =\Sigma _{k=2} ^{\infty} a_{k-1}

Well, the question is, does the theorem permit all these?

 

[rbj]

For the series such that: \Sigma _{n=1} ^{\infty} a_n =\Sigma _{n=1} ^{\infty} b_n A certain theorem says that these series are equal even if a_n = b_n only for n>m. That is, even if two infinite series differ for a finite number of terms, it will still converge for the same sum.

well, that "certain theorem" is not true. pretty obviously not true.

 

[rbj]

well, that "certain theorem" is not true. pretty obviously not true.

perhaps i should ask you to restate the certain theorem, but from i am surmising from what you stated, i can't think of a close variation of your words that would add up to a theorem that is true.

 

[mathsciguy]

Sorry, I might not remember it very well, I'll just do a direct quote from a textbook then.
Woah, I now realize I'm completely wrong, my 'theorem' up there was a product of my cluttered head, so please excuse it. Here goes the correct theorem:

If \Sigma _{n=1} ^{\infty} a_n and \Sigma _{n=1} ^{\infty} b_n are two infinite series, differing only in their first m terms (i.e., a_k=b_k if k>m, then either both series converge or both series diverge.

Well, my question now is, suppose \Sigma _{n=1} ^{\infty} a_k is a series that converges to some sum S. I was thinking that the series \Sigma _{n=2} ^{\infty} a_{n-1} also converges to the same sum S, or even \Sigma _{n=m} ^{\infty} a_n for some other integer m (even for negative integers) is also S. Does the theorem permit this kind of manipulation? Well, the theorem says nothing about the value of the sum, only about its convergence, would anybody expound on this idea?

 

[micromass]

Sorry, I might not remember it very well, I'll just do a direct quote from a textbook then.
Woah, I now realize I'm completely wrong, my 'theorem' up there was a product of my cluttered head, so please excuse it. Here goes the correct theorem:

If \Sigma _{n=1} ^{\infty} a_n and \Sigma _{n=1} ^{\infty} b_n are two infinite series, differing only in their first m terms (i.e., a_k=b_k if k>m, then either both series converge or both series diverge.

Well, my question now is, suppose \Sigma _{n=1} ^{\infty} a_k is a series that converges to some sum S. I was thinking that the series \Sigma _{n=2} ^{\infty} a_{n-1} also converges to the same sum S

Yes, this series converges to the same sum. But that doesn't follow from the theorem. Rather it follows because the terms of the series are equal.

or even \Sigma _{n=m} ^{\infty} a_n for some other integer m (even for negative integers) is also S.

The theorem tells you that this converges. But it tells you nothing about the sum.

 

[mathsciguy]

Ah, yes, yes, now that I have completely rewritten the theorem, and with your clarification I now understand better.

I sometimes feel like I'm hunting Easter eggs when I'm looking at theorems.

 

[rbj]

If \Sigma _{n=1} ^{\infty} a_n and \Sigma _{n=1} ^{\infty} b_n are two infinite series, differing only in their first m terms (i.e., a_k=b_k if k>m, then either both series converge or both series diverge.

when two different series both converge, they may converge to two different finite numbers.

 

[mathsciguy]

Sorry for necroing an old thread, but I've gone back, and am reviewing my infinite series knowledge. I thought the previous discussions here might be useful for my query.

My question is, can we generalize an infinite series so that it can be denoted by:
\Sigma ^{\infty} _{n=m}a_n where m is any integer? I'm trying to ask if we can shift the starting index to any integer.

Because, from the textbook definition of an infinite series, it is defined as:

If \{u_n\} is a sequence and
s_n=u_1+u_2+u_3+...+u_n
then \{s_n\} is a sequence of partial sums called an infinite series denoted by
\Sigma ^{\infty} _{n=1}u_n=u_1+u_2+u_3+...+u_n+...
The numbers u_1,u_2,...,u_n,... are the terms of the infinite series.

From this definition, there is an associated sequence of partial sums to an infinite series, but a sequence is only defined to have a positive integral domain. Does that mean, for an infinite series, the lower bound of the starting index is 1?

 

[mathsciguy]

I'd really appreciate it if someone would reply. Is my question too trivial?

 

[micromass]

Sorry for necroing an old thread, but I've gone back, and am reviewing my infinite series knowledge. I thought the previous discussions here might be useful for my query.

My question is, can we generalize an infinite series so that it can be denoted by:
\Sigma ^{\infty} _{n=m}a_n where m is any integer? I'm trying to ask if we can shift the starting index to any integer.

Because, from the textbook definition of an infinite series, it is defined as:

If \{u_n\} is a sequence and
s_n=u_1+u_2+u_3+...+u_n
then \{s_n\} is a sequence of partial sums called an infinite series denoted by
\Sigma ^{\infty} _{n=1}u_n=u_1+u_2+u_3+...+u_n+...
The numbers u_1,u_2,...,u_n,... are the terms of the infinite series.

From this definition, there is an associated sequence of partial sums to an infinite series, but a sequence is only defined to have a positive integral domain. Does that mean, for an infinite series, the lower bound of the starting index is 1?

Sure, you can generalize the definition so that it starts from any integer instead of the integer 1.

 

[mathsciguy]

Sure, you can generalize the definition so that it starts from any integer instead of the integer 1.

Hm, does that mean, the sequence of partial sums of the said series will also have 'm' as its starting index?

 

[Office_Shredder]

Hm, does that mean, the sequence of partial sums of the said series will also have 'm' as its starting index?

Yup, but doing proofs with that is going to be annoying (lots of indexing that needs to match up). It would be easier to just say if your sum starts at n=4 say, then a₁, a₂, and a₃ are equal to zero