Proof by contradicion Sequences

[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?

 

[sutupidmath]

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:

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

Which is clearly a contradiction.


Will this work?

 

[matticus]

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

 

[sutupidmath]

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

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.

 

[matticus]

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)

 

[Gib Z]

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.

 

[sutupidmath]

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.

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!

 

[MathematicalPhysicist]

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.

 

[sutupidmath]

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.

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.

 

[MathematicalPhysicist]

I shoud relook my notes of first year if I haven't 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.

 

[JG89]

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 $$\lim_{n \rightarrow \infty} y_n = 0$$.

 

[sutupidmath]

This is how I would tackle the problem.


$$\lim_{n \rightarrow \infty} \frac{y_n_+_1}{y_n} = \lim_{n \rightarrow \infty} y_n_+_1 * \lim_{n \rightarrow \infty} \frac{1}{y_n} = 0$$.

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.

 

[JG89]

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.

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. "