Does Convergence of a Series Imply that the Limit of n Times the Term is Zero?

In summary, if the series of numbers a_n converges, then the limit of a_n as n approaches infinity is 0.
  • #1
Ed Quanta
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I know that if the series of (a)n (n is a subscript) converges, then the lim (a)n=0. How can I show that if the series of (a)n converges, then lim n(a)n=0?

Or rather if a1 +a2 +a3 +...+an=0, then lim n*(a)n=0?

Not sure how to show this, but I know the proof involves the cauchy criterion for series. Help anyone?
 
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  • #2
Hint:
Let:
[tex]S_n=\sum_{k=1}^na_k[/tex]

Notice that
[tex]a_n=S_n-S_{n-1}[/tex]
 
  • #3
I am sorry but can I ask you for another hint. I understand that what you wrote is true, but what am I supposed to do with it. Using the Caucy Criterion for series, I know that there is an N such that for all n>m>N,
sn-sm< for all epsilon >0. But where do I go from here? Sorry for my slowness in comprehension.
 
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  • #4
Oh, actually my hint goes with a proof that doesn't use the Cauchy criterion.
Just assume the series is convergent and take the limit on both sides of the equation.

Ok, so the Cauchy criterion is:
A series is a Cauchy-series if for every [itex]\epsilon>0[/itex] there is a N>0, such that [itex]|S_n-S_m|<\epsilon[/itex] voor any n,m>N.

In particular, it holds for m=n+1.
Now write out [itex]|S_n-S_m|[/itex] and see what you get for m=n+1.
 
  • #5
So this is sort of an inductive proof? We show the inequality holds for n=m+1, and since the an terms are non increasing, we know Sn-Sm<epsilon will hold where n>m+1. Correct?
 
  • #6
Ed Quanta said:
So this is sort of an inductive proof? We show the inequality holds for n=m+1, and since the an terms are non increasing, we know Sn-Sm<epsilon will hold where n>m+1. Correct?
NO! It's not an inductive proof.
We have to show that IF a series [itex]\sum a_n[/itex] converges, then [itex]\lim_{n\rightarrow \infty} a_n= 0[/itex].

So we ASSUME our series is convergent, then we know that for any [itex]\epsilon >0[/itex] we can find an N such that [itex]|Sn-Sm|<\epsilon[/itex] for all n,m>N.

From the above assumption we have to show that:
for any [itex]\epsilon>0[/itex] there exists an N, such that [itex]|a_n|<\epsilon[/itex] whenever n>N.
(This is just the definition of [itex]\lim_{n\rightarrow \infty} a_n= 0[/itex].)

What I meant was. If you plug m=n+1 in |Sm-Sn| and write it in terms of sums, then...

(it's still a hint).
 
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  • #7
Galileo said:
NO! It's not an inductive proof.
We have to show that IF a series [itex]\sum a_n[/itex] converges, then [itex]\lim_{n\rightarrow \infty} a_n= 0[/itex].

No, that wasn't the original question.

The problem was to prove that "If [itex]\Sigma a_n[/itex] converges, then
[itex]lim_{n\rightarrow \infty}na_n= 0[/itex].
 

FAQ: Does Convergence of a Series Imply that the Limit of n Times the Term is Zero?

What is the Cauchy Criterion for Series?

The Cauchy Criterion for Series is a mathematical test used to determine if an infinite series converges or diverges. It states that if the terms of a series become arbitrarily close to each other as the series progresses, then the series will converge.

How is the Cauchy Criterion used to determine convergence?

To use the Cauchy Criterion, we look at the terms of the series and check if they become arbitrarily close to each other as the series progresses. If this is the case, then the series is said to converge. If the terms do not become close to each other, then the series diverges.

What is the mathematical notation for the Cauchy Criterion?

The Cauchy Criterion is often written in mathematical notation as: For every positive number ε, there exists a positive integer N such that for all m and n greater than N, the absolute value of the difference between the m-th and n-th terms of the series is less than ε.

Can the Cauchy Criterion be used to determine if a series diverges?

Yes, the Cauchy Criterion can be used to determine both convergence and divergence of a series. If the terms of the series do not become arbitrarily close to each other, then the series will diverge.

How is the Cauchy Criterion related to the concept of limit?

The Cauchy Criterion is closely related to the concept of limit. In fact, the Cauchy Criterion is essentially a restatement of the limit definition for a sequence. It states that if the terms of a series approach a limit as the series progresses, then the series will converge.

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