# Cauchy Criterion for Series

1. Nov 8, 2004

### Ed Quanta

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?

2. Nov 8, 2004

### Galileo

Hint:
Let:
$$S_n=\sum_{k=1}^na_k$$

Notice that
$$a_n=S_n-S_{n-1}$$

3. Nov 8, 2004

### Ed Quanta

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.

Last edited: Nov 8, 2004
4. Nov 8, 2004

### Galileo

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 $\epsilon>0$ there is a N>0, such that $|S_n-S_m|<\epsilon$ voor any n,m>N.

In particular, it holds for m=n+1.
Now write out $|S_n-S_m|$ and see what you get for m=n+1.

5. Nov 8, 2004

### Ed Quanta

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. Nov 8, 2004

### Galileo

NO! It's not an inductive proof.
We have to show that IF a series $\sum a_n$ converges, then $\lim_{n\rightarrow \infty} a_n= 0$.

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

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

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).

Last edited: Nov 8, 2004
7. Nov 9, 2004

### HallsofIvy

No, that wasn't the original question.

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