Define the notation used here in describing a series?

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SUMMARY

The discussion centers on defining the notation for sequences and series in the context of the infinite series ∑∞1 ((1/n) - 1/(n+1)). The terms an, Sn, and Rn are clarified, where an represents the nth term of the series, Sn denotes the nth partial sum, and Rn is the remainder after n terms, defined as Rn = S - Sn, where S is the limit of Sn as n approaches infinity. The distinction between sequences and series is emphasized, with sequences referring to the individual terms and series referring to their sums.

PREREQUISITES
  • Understanding of infinite series and convergence
  • Familiarity with sequences and summation notation
  • Basic knowledge of limits in calculus
  • Ability to manipulate algebraic expressions involving series
NEXT STEPS
  • Study the properties of convergent and divergent series
  • Learn about the concept of partial sums and their significance
  • Explore the use of the remainder term Rn in estimating errors in series
  • Investigate specific examples of series, such as geometric and harmonic series
USEFUL FOR

Students of calculus, mathematics educators, and anyone seeking to deepen their understanding of sequences and series in mathematical analysis.

whitejac
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Homework Statement


"For the given series, write formulas for the sequences an , Sn, Rn and find the limit as n->∞ (if it exists)

Homework Equations


1 ((1/n) - 1/(n+1)

The Attempt at a Solution


I know how to take the limit, that's no problem. I'm a bit confused about what an , Sn, Rn are referring to. Maybe I'm just misunderstanding the usage...

My book speaks generally of defining them (by that i mean, no formal definition is given but it often refers to a previous example) however it does say this:

Let us call the terms of the series a so that the series
is
"a1 + a2 + a3 + a4 + · · · + ann + · · · ."
Remember that the three dots mean that there is never a last term; the series goes
on without end. Now consider the sums Sn that we obtain by adding more and
more terms of the series. We define
S1 = a1,
S2 = a1 + a2,
S3 = a1 + a2 + a3,
· · ·
Sn = a1 + a2 + a3 + · · · + an


It then goes further to say:

The difference Rn = S − Sn is called the remainder (or the remainder after n
terms). From (4.6), we see that...

Where S is the lim n-> ∞. So can Rn not be found if there's no limit?
Also, why even write down an if it's just the same as Sn?
 
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whitejac said:
So can ##R_n## not be found if there's no limit?
That is correct.
why even write down ##a_n## if it's just the same as ##S_n##?
They are not the same. ##S_n## is the sum of ##a_1## to ##a_n##, ie
$$S_n\triangleq \sum_{i=1}^n a_i$$
Also, in words sometimes a distinction is made by using the word 'sequence' to refer to the ##a_n##s and 'series' to refer to the ##S_n##s. It's good to bear that in mind to head off potential confusion if they use that terminology.
 
whitejac said:

Homework Statement


"For the given series, write formulas for the sequences an , Sn, Rn and find the limit as n->∞ (if it exists)

Homework Equations


1 ((1/n) - 1/(n+1)

The Attempt at a Solution


I know how to take the limit, that's no problem. I'm a bit confused about what an , Sn, Rn are referring to. Maybe I'm just misunderstanding the usage...

My book speaks generally of defining them (by that i mean, no formal definition is given but it often refers to a previous example) however it does say this:

Let us call the terms of the series a so that the series
is
"a1 + a2 + a3 + a4 + · · · + ann + · · · ."
Remember that the three dots mean that there is never a last term; the series goes
on without end. Now consider the sums Sn that we obtain by adding more and
more terms of the series. We define
S1 = a1,
S2 = a1 + a2,
S3 = a1 + a2 + a3,
· · ·
Sn = a1 + a2 + a3 + · · · + an


It then goes further to say:

The difference Rn = S − Sn is called the remainder (or the remainder after n
terms). From (4.6), we see that...

Where S is the lim n-> ∞. So can Rn not be found if there's no limit?
Also, why even write down an if it's just the same as Sn?

Who says that ##a_n## is the same as ##S_n##? YOU wrote
$$S_n = a_1 + a_2 + \cdots + a_n$$
##S_n## and ##a_n## look a lot different to me.

Anyway, if ##S = \lim_{n \to \infty} S_n## then it makes sense to speak of the difference between ##S## and ##S_n##; that difference is called ##R_n##. In other words, ##S = S_n + R_n##.

In some (rare) cases we can say exactly what is ##R_n## (for example, for a geometric series), but usually we have to be content with producing bounds on the value of ##R_n## instead of an exact value. Still, that is often very useful, especially when we want to evaluate an infinite series numerically by stopping at ##S_n## for some finite ##n##. A bound on ##R_n## will tell us an upper bound on the error we would be making by stopping at ##n##.
 
You need to explain what "the given series" is.

From the information you quoted, apparently ##a_n## is the nth term of a series.

Is the "given series" the series whose ##n##-th term is ##a_n = \frac{1}{n} - \frac{1}{n+1} ## ?##S_n## is the "##n##-th partial sum" of the series. ##S_n = \sum_{i=1}^n a_i ##, so ##S_n## is not the same thing as ##a_n##.##S## is the sum of the series ## S = \lim_{n \rightarrow \infty} S_n ##

##R_n## is ## S - S_n ## You are correct that ##R## is undefined if ##S## doesn't exist.
 
Stephen Tashi said:
You need to explain what "the given series" is.

From the information you quoted, apparently ##a_n## is the nth term of a series.

Is the "given series" the series whose ##n##-th term is ##a_n = \frac{1}{n} - \frac{1}{n+1} ## ?##S_n## is the "##n##-th partial sum" of the series. ##S_n = \sum_{i=1}^n a_i ##, so ##S_n## is not the same thing as ##a_n##.##S## is the sum of the series ## S = \lim_{n \rightarrow \infty} S_n ##

##R_n## is ## S - S_n ## You are correct that ##R## is undefined if ##S## doesn't exist.
How can you say that an is a term if the question requests a formula?
and I apologize, the given series is the series found in section two under "useful formulas,"
whitejac said:
∑∞1 ((1/n) - 1/(n+1)
This is the series with which I am supposed to be determining the formulas "an, Sn, Rn."
If an is the nth term in the series, then that would just be the given equation, right?
if n = 27, then the nth term would be
((1/27) - (1/28))
Then Sn would be the partial sum to the nth term, which would be
(1/1) - (1/2) + (1/2) - (1/3) + (1/3) - 1/4 ... + (1/27) - (1/28) = 1 - (1/28) = (27/28)
am i understanding the definitions correctly?
 
whitejac said:
How can you say that an is a term if the question requests a formula?
and I apologize, the given series is the series found in section two under "useful formulas,"

This is the series with which I am supposed to be determining the formulas "an, Sn, Rn."
If an is the nth term in the series, then that would just be the given equation, right?
if n = 27, then the nth term would be
((1/27) - (1/28))
Then Sn would be the partial sum to the nth term, which would be
(1/1) - (1/2) + (1/2) - (1/3) + (1/3) - 1/4 ... + (1/27) - (1/28) = 1 - (1/28) = (27/28)
am i understanding the definitions correctly?

Almost. That last line is the sum to the ##27##th term ##S_{27}##, not to the ##n##th term ##S_n##.
 
whitejac said:
How can you say that an is a term if the question requests a formula?
write formulas for the sequences an , Sn, Rn
In other words, from the given summation (which is NOT an equation), write expressions for each of the sequences listed above.
whitejac said:
and I apologize, the given series is the series found in section two under "useful formulas,"

This is the series with which I am supposed to be determining the formulas "an, Sn, Rn."
If an is the nth term in the series, then that would just be the given equation, right?
No, and again, the summation you showed is not an equation. An equation has this symbol in it = .
whitejac said:
if n = 27, then the nth term would be
((1/27) - (1/28))
Then Sn would be the partial sum to the nth term, which would be
(1/1) - (1/2) + (1/2) - (1/3) + (1/3) - 1/4 ... + (1/27) - (1/28) = 1 - (1/28) = (27/28)
am i understanding the definitions correctly?

When you're working with infinite series, such as ##\sum_{j = 1}^{\infty} \frac 1 j##, there are several bits of terminology that beginning students have trouble with.

A sequence is basically a list of numbers. A series is the sum of a possibly infinite number of terms of the related sequence of terms.

With an infinite series, you have the following:
  • There is the underlying sequence of terms, ##\{a_1, a_2, a_3, \dots \}##, which in my example is the sequence {1/1, 1/2, 1/3, 1/4, ...}
  • There is the sequence of partial sums, ##\{S_1, S_2, S_3, \dots \}##, which in my example is the sequence {1, 3/2, 11/6, ...}. S1 is the sum of the first term in the sequence of terms (IOW, it's the first term). S2 is the sum of the first two terms in the sequence of terms, S3 is the sum of the first three terms, and so on.
  • There is the series itself. A series converges if and only if its sequence of partial sums converges.
 
Last edited:
whitejac said:
How can you say that an is a term if the question requests a formula?

I don't see any difficulty in giving a formula as the answer for a "term". The n-th term is given by a function of n.

If an is the nth term in the series, then that would just be the given equation, right?

There is no given "equation". (There is no "=" sign in the given expression ##\sum_{1}^\infty (\frac{1}{n} - \frac{1}{n+1} ) ## .) The given information is "an expression" showing a summation that involves a function of n. Yes, ##a_n## is the given function.

Most text would have written the given sum as ##\sum_{n=1}^\infty (\frac{1}{n} - \frac{1}{n+1} )## Did your text leave out the "n = 1" ?

if n = 27, then the nth term would be
((1/27) - (1/28))

Yes.

Then Sn would be the partial sum to the nth term, which would be
(1/1) - (1/2) + (1/2) - (1/3) + (1/3) - 1/4 ... + (1/27) - (1/28) = 1 - (1/28) = (27/28)
Yes, with the understanding that n = 27.
am i understanding the definitions correctly?

I think so.
 
Hmm, I think I understand now. The definitions and nomenclature definitely slips me up. Thank you all for your patience!
 

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