Proving the Convergence of a Series: A Simple and Concise Guide

In summary, the series \sum_{n=m}^\infty a_n converges if and only if m is an integer such that the series \sum_{n=m}^\infty a_n converges.
  • #1
synkk
216
0
show that if ## a_n ## is any sequence and m is any integer such that the series ## \displaystyle \sum_{n=m}^\infty a_n ## converges, then ## \displaystyle\lim_{k \to \infty} \sum_{n=k}^{\infty
} a_n = 0 ##

let ## S_N = \displaystyle \sum_{n=m}^N a_n ##

## \lim_{N \to \infty} S_N = l ## from definition
also,
## \lim_{N \to \infty} S_{N-1} = l ##

## S_N - S_{N-1} = a_N ##

## \lim_{N \to \infty} (S_N - S_{N-1}) = \lim_{N \to \infty}(S_N) - \lim_{N \to \infty} S_{N-1} = 0 ## then ## \lim_{N \to \infty} a_N = 0 ## i.e. ## \lim_{N \to \infty} \sum_{n=m}^N a_n = 0 \Rightarrow \lim_{k \to \infty} \sum_{n=k}^{\infty} a_n = 0 ##

as you can see I don't really know how to prove this statement, but I have attempted it.

Could someone show me how I can proceed to do the proof properly?
 
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  • #2
synkk said:
show that if ## a_n ## is any sequence and m is any integer such that the series ## \displaystyle \sum_{n=m}^\infty a_n ## converges, then ## \displaystyle\lim_{k \to \infty} \sum_{n=k}^{\infty
} a_n = 0 ##

let ## S_N = \displaystyle \sum_{n=m}^N a_n ##

## \lim_{N \to \infty} S_N = l ## from definition
also,
## \lim_{N \to \infty} S_{N-1} = l ##

## S_N - S_{N-1} = a_N ##

## \lim_{N \to \infty} (S_N - S_{N-1}) = \lim_{N \to \infty}(S_N) - \lim_{N \to \infty} S_{N-1} = 0 ## then ## \lim_{N \to \infty} a_N = 0 ## i.e. ## \lim_{N \to \infty} \sum_{n=m}^N a_n = 0 \Rightarrow \lim_{k \to \infty} \sum_{n=k}^{\infty} a_n = 0 ##

as you can see I don't really know how to prove this statement, but I have attempted it.

Could someone show me how I can proceed to do the proof properly?

You are given that
[tex]
\sum_{n=m}^\infty a_n
[/tex]
converges. Thus you have
[tex]
L = \sum_{n=m}^\infty a_n = S_{k-1} + \sum_{n = k}^{\infty} a_n
[/tex]
so that
[tex]
|L - S_{k-1}| = \left| \sum_{n = k}^{\infty} a_n \right|
[/tex]
 
  • #3
pasmith said:
You are given that
[tex]
\sum_{n=m}^\infty a_n
[/tex]
converges. Thus you have
[tex]
L = \sum_{n=m}^\infty a_n = S_{k-1} + \sum_{n = k}^{\infty} a_n
[/tex]
so that
[tex]
|L - S_{k-1}| = \left| \sum_{n = k}^{\infty} a_n \right|
[/tex]
how did you get ## L = \sum_{n=m}^\infty a_n = S_{k-1} + \sum_{n=k}^{\infty} a_n ##?
 
  • #4
synkk said:
how did you get ## L = \sum_{n=m}^\infty a_n = S_{k-1} + \sum_{n=k}^{\infty} a_n ##?

They've just written it as a sum of two series.
 
  • #5
I do not believe it is correct though.
It should read [itex]L=\sum_{n=0}^\infty a_n = S_{k-1}+\sum_{n=k}^\infty a_n[/itex].
This is to be seen from the definition of [itex]S_k = \sum_{n=0}^k a_n[/itex].

The final step in post #2 is correct. (Although I'm not entirely sure why the absolute value shows up, but it doesn't make any difference here)
 
  • #6
JorisL said:
I do not believe it is correct though.
It should read [itex]L=\sum_{n=0}^\infty a_n = S_{k-1}+\sum_{n=k}^\infty a_n[/itex].
This is to be seen from the definition of [itex]S_k = \sum_{n=0}^k a_n[/itex].

Read the OP. We are told that [itex]\sum_{n=m}^\infty a_n[/itex] converges for some integer [itex]m[/itex]. In particular, we are not told that [itex]m[/itex] is positive. Therefore the OP's definition of [itex]S_k = \sum_{n=m}^{k} a_n[/itex] is correct.

However, in any event the behaviour of a finite number of terms at the beginning of a sequence does not affect convergence of the series.
 
  • #7
My bad, sorry for the confusion.
But this doesn't change anything indeed.
 

1. How do I know if a series is convergent?

A series is convergent if the sequence of partial sums approaches a finite limit as the number of terms increases. This can be determined using various convergence tests, such as the comparison test, ratio test, or integral test.

2. What is the difference between absolute and conditional convergence?

Absolute convergence means that the series converges regardless of the order in which the terms are added, while conditional convergence means that the series only converges if the terms are added in a specific order.

3. Can a divergent series be proven to be convergent?

No, a series that is divergent cannot be proven to be convergent. However, if a series fails the divergence test, it can still be proven to be divergent.

4. What is the role of the limit comparison test in proving convergence?

The limit comparison test is used to compare the convergence of a given series to a known convergent or divergent series. This can be helpful in determining the convergence of a series that is difficult to evaluate using other tests.

5. Can a series be both convergent and divergent?

No, a series can only be either convergent or divergent. If the series fails any of the convergence tests, it is considered to be divergent.

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