1. Jan 8, 2009

### sutupidmath

Well, implicitly in a problem i came accros something that looks like it first requires to establish the following result:(to be more precise, the author uses the following result in the problem)

Let $$\{y_n\}$$ be a sequence. If

$$\lim_{n\rightarrow \infty}\frac{y_{n+1}}{y_n}=0,=> \lim_{n\rightarrow \infty}y_n=0$$

I tried a proof by contradicion, but i couln't really prove it.

SO, any hints?

2. Jan 8, 2009

### sutupidmath

Re: Sequences!

Ok,here it is a thought, i'm not sure whether this would work though.

Suppose that, $$lim_{n\rightarrow\infty}y_n=a$$ where a is different from zero.

Then, we know that $$lim_{n\rightarrow\infty}y_{n+1}=a$$ as well.

So, now we would have:

$$\lim_{n\rightarrow\infty}\frac{y_{n+1}}{y_n}=\frac{\lim_{n\rightarrow\infty}y_{n+1}}{lim_{n\rightarrow\infty}y_n}=\frac{a}{a}=1$$

Will this work?

3. Jan 8, 2009

### matticus

Re: Sequences!

you still leave the possibility that y(n) doesn't converge at all.

4. Jan 8, 2009

### sutupidmath

Re: Sequences!

Well, that's not a problem. I forgot to mention, that i can show that y_n is a monotonic and bounded sequence, hence it must converge.

5. Jan 8, 2009

### matticus

Re: Sequences!

right. yes, the proof is valid. (just to clarify the step, lim y(n+1) converges since y(n) converges and hence every subsequence must converge to the same limit)

6. Jan 10, 2009

### Gib Z

Re: Sequences!

Just a thought, but if you are allowed to recognize the first limit as a positive result for the ratio convergence test, then the sum of the sequence must converge and hence the terms must approach zero as required.

7. Jan 10, 2009

### sutupidmath

Re: Sequences!

Well, that would work also in the case where y_n would be greater than zero for some n>N, and also in the case of alternating series, since in that case we could use the ratio test for alternating series:

$$\lim_{n\rightarrow \infty}\left|\frac{y_{n+1}}{y_n}\right|=0<1$$ so the corresponding series

$$\sum y_n$$ would converge absolutely and thus would be convergent as well, so y_n-->0 as n-->infinity.

(p.s. i was basically aiming for a proof by contradiction in my first post)

Cheers!

8. Jan 10, 2009

### MathematicalPhysicist

Re: Sequences!

It's really rather simple when you think of it.
let's go with definitions:

lim (y_n/y_n-1)=L iff for every e>0 there exists N s.t for every n>N -e<y_n/y_n-1-L<e
L-e<y_n+1/y_n<L+e
we assume that y_n doesn't vanish for each n.
y_n=(y_n/y_n-1)*(y_n-1/y_n-2)*....*(y_2/y_1)*y_1 the ratio sequence converges to L thus it is bounded let M be this bound, assume that n-k=N for some k natural, then y_n-L=(y_n/y_n-1)*(y_n-1/y_n-2)*....*(y_2/y_1)*y_1-L

y_n<=y_1*M^k(y_n/y_n-1)...(y_n-k+1/y_n-k)<y_1*M^k(L+e)^(n-k) as n->infinity y_n->0.

Did I do a mess with this question, or am I on the right track, I had this in my first year at university so I should remember.

9. Jan 10, 2009

### sutupidmath

Re: Sequences!

Hmm..well, you are saying on this line that y_n is bounded by:

$$y_n<y_1*M^k(L+\epsilon)^{n-k}$$

but, if n-->infinity, how can we be sure that

$$(L+\epsilon)^{n-k}-->0$$ since obviously M^k doesn't go to zero.

unless, $$|L+\epsilon|<1$$ of which is not necessarly true in general.

I am probbably missing something in this whole picture, since i didn't analyze it a lot.

10. Jan 10, 2009

### MathematicalPhysicist

Re: Sequences!

I shoud relook my notes of first year if I havent thrown them away, I remember asking my TA this question, and it was also in courant and fritz text.
There is some trick in it that I forgot, perhaps someone here remembers the proof that goes via epsilon.

11. Jan 11, 2009

### JG89

Re: Sequences!

This is how I would tackle the problem.

$$\lim_{n \rightarrow \infty} \frac{y_n_+_1}{y_n} = \lim_{n \rightarrow \infty} y_n_+_1 * \lim_{n \rightarrow \infty} \frac{1}{y_n} = 0$$, implying that either of these limits converge to 0. If $$\lim_{n \rightarrow \infty} y_n_+_1 = 0$$ then $$\lim_{n \rightarrow \infty} y_n = 0$$ and we're done. So we take a look at the other limit. Suppose that $$\lim_{n \rightarrow \infty} \frac{1}{y_n} = 0$$. Then $$1 = 0*\lim_{n \rightarrow \infty} y_n$$. Since $$y_n$$ converges, say to "a", then we would have 1 = 0*a = 0, an obvious contradiction. Thus $$\lim_{n \rightarrow \infty} y_n_+_1 = 0$$ implying that $$\lim_{n \rightarrow \infty} y_n = 0$$.

12. Jan 11, 2009

### sutupidmath

Re: Sequences!

I don't think you can separate the limits this way at first place, since if you are doing so, you are assuming that the limit of y_n as n goes to infinity exists, of which we are not sure, and we have to prove as well. So, i don't think this would work.

13. Jan 11, 2009

### JG89

Re: Sequences!

I was thinking about that when I first thought of it, but then I saw you write this "I forgot to mention, that i can show that y_n is a monotonic and bounded sequence, hence it must converge. "