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Show sequence is cauchy

by Unassuming
Tags: cauchy, sequence
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Unassuming
#1
Oct16-08, 08:57 PM
P: 169
For n in the naturals, let

[tex]p_n = 1 + \frac{1}{2!} + ... + \frac{1}{n!}[/tex]

Show it is cauchy.


Attempt:

I have set up |p_n+k - p_n | < e , and I have solved for this.

I got [tex]|p_{n+k} - p_n | = \frac{1}{(n+1)!} + ... + \frac{1}{(n+k)!}[/tex]

I am trying to follow an example in the book. I now need to find a telescoping sequence that is a little bit greater than my sequence above. It should also contain what seems like two fractions, or parts, with one of them converging and the other adding an arbitrary "k". I would appreciate any hints on this.

I got stuck on how to break up, 1 / (n+k-1)! . I am not sure if that is worthwhile or I am totally off.
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Mark44
#2
Oct16-08, 09:52 PM
Mentor
P: 21,286
Quote Quote by Unassuming View Post
For n in the naturals, let

[tex]p_n = 1 + \frac{1}{2!} + ... + \frac{1}{n!}[/tex]

Show it is cauchy.


Attempt:

I have set up |p_n+k - p_n | < e , and I have solved for this.

I got [tex]|p_{n+k} - p_n | = \frac{1}{(n+1)!} + ... + \frac{1}{(n+k)!}[/tex]

I am trying to follow an example in the book. I now need to find a telescoping sequence that is a little bit greater than my sequence above. It should also contain what seems like two fractions, or parts, with one of them converging and the other adding an arbitrary "k". I would appreciate any hints on this.

I got stuck on how to break up, 1 / (n+k-1)! . I am not sure if that is worthwhile or I am totally off.
You have k terms, the largest of which is 1/(n + 1)! Is that enough of a hint?
Unassuming
#3
Oct16-08, 11:25 PM
P: 169
Your hint helped but since this is my first cauchy problem I am still lost. I am getting lost when I try to find the partial fractions of

[tex]\frac{1}{n!(n+k)}[/tex]

I am doing this in order to get the telescoping sequence.

Mark44
#4
Oct16-08, 11:48 PM
Mentor
P: 21,286
Show sequence is cauchy

Quote Quote by Unassuming View Post
Your hint helped but since this is my first cauchy problem I am still lost. I am getting lost when I try to find the partial fractions of

[tex]\frac{1}{n!(n+k)}[/tex]

I am doing this in order to get the telescoping sequence.
So why do you think you need to decompose the fraction and why do you think you need a telescoping sequence? Given a positive number epsilon, all you need to do is find a number N so that for all m and n larger than N, any two terms in your sequence are closer together than epsilon.

HINT: You have k terms (count 'em!) on the right, the largest of which is 1/(n + 1)!
boombaby
#5
Oct17-08, 03:31 AM
P: 130
I'm not sure how to use your hint, Mark44....
are you going to prove it by saying that
[tex]\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...+\frac{1}{(n+k)!}<\frac{k}{(n+1)!}[/tex],
which looks as if it approaches to 0 as N goes to infinity. So it can be made less than a given epsilon?
But, I realize that k is independent on the given epsilon. So if you find a N. I can choose k large enough, say, k=(n+1)! which will make that inequality useless.
So how can it be done?

A different hint: prove its convergenece as an upper bounded increasing sequences, which implies also it's a cauchy.

PS. Well, after a second thought, The hint of Mark44 really works, but a bit tricky
Unassuming
#6
Oct17-08, 03:48 PM
P: 169
Quote Quote by boombaby View Post
[tex]\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...+\frac{1}{(n+k)!}<\frac{k}{(n+1)!}[/tex],
I got [tex] \frac{k}{(n+1)!} = \frac{k}{n!} - \frac{kn}{(n+1)!} [/tex]

I can't get anywhere. I want to say something like k/n! is convergent, therefore it is cauchy and we can say that,

[tex]| \frac{k}{n!} - \frac{kn}{(n+1)!} | < e [/tex]

I know that can't work because there is an n in the numerator but what else can I do?
Unassuming
#7
Oct17-08, 07:47 PM
P: 169
Okay, I feel good. Somebody shoot me down!

Since 1/n! converges to 0 and n->inf , we can let e>0 and say there exists N in Naturals such that for all n >= N,

| 1/n! - 0 | < e/k.

Therefore,

| s_n+k - s_n | = 1/(n+1)! + ... + 1/(n+k)! < k/ (n+1)! < k/n! < k(e/k) = e.

Please, please, please let this be right. I am ready to move on.
boombaby
#8
Oct17-08, 08:17 PM
P: 130
this is not right.. as I said before, your k should be independent on epsilon.
try
[tex]\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...+\frac{1}{(n+k)!}=\frac{1}{(n+1)!} (1+\frac{1}{(n+2)}+\frac{1}{(n+2)(n+3)}+...\frac{1}{(n+2)...(n+k)})<\fr ac{1}{(n+1)!} (1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\frac{1}{(n+1)^{k}})<\frac{1}{n! n}[/tex]
so we can choose a N large enough, no matter how large your k is, its sum is less than a given epislon.
dirk_mec1
#9
Oct18-08, 06:28 AM
P: 682
[tex]
\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...+\frac{1}{(n+ k)!}=\frac{1}{(n+1)!} (1+\frac{1}{(n+2)}+\frac{1}{(n+2)(n+3)}+...\frac{1 }{(n+2)...(n+k)})<\frac{1}{(n+1)!} (1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\frac{1}{(n +1)^{k}})<
[/tex]



[tex] < \frac{1}{(n+1)!} \sum_{j=0}^{\infty} \frac{1}{(n+1)^j} < \sum_{j=0}^{\infty} \frac{1}{(n+1)^j} = \frac{n+1}{n} [/tex]
boombaby
#10
Oct20-08, 02:57 AM
P: 130
no...it is useless that you have "........< (n+1)/n"
Unassuming
#11
Oct20-08, 08:43 AM
P: 169
Choose N> 1/e. Then for all n > N we have that,

[tex]

\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...+\frac{1}{(n+k)!}=\frac{1}{(n+1)!} (1+\frac{1}{(n+2)}+\frac{1}{(n+2)(n+3)}+...\frac{1}{(n+2)...(n+k)})<\fr ac{1}{(n+1)!} (1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\frac{1}{(n+1)^{k}})<\frac{1}{n! n}<\frac{1}{n}<\varepsilon

[/tex]

Therefore, [tex]|s_{n+k}-s_n|<\varepsilon[/tex].

Is that right?

EDIT: Is that LaTeX showing up for anybody else?
dirk_mec1
#12
Oct20-08, 11:29 AM
P: 682
[tex]
\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...+\frac{1}{(n+ k)!}=\frac{1}{(n+1)!} (1+\frac{1}{(n+2)}+\frac{1}{(n+2)(n+3)}+...\frac{1 }{(n+2)...(n+k)})<\frac{1}{(n+1)!} (1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\frac{1}{(n +1)^{k}})<\frac{1}{n!n}<\frac{1}{n}<\varepsilon
[/tex]

\\Edit: indeed latex isn't working
boombaby
#13
Oct20-08, 12:49 PM
P: 130
test:
[tex]\frac{1}{n!} \int^{1}_{0} sin x dx[/tex]

Edit: indeed
dirk_mec1
#14
Oct22-08, 03:43 AM
P: 682
[tex]

\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...+\frac{1}{(n+ k)!}=\frac{1}{(n+1)!} (1+\frac{1}{(n+2)}+\frac{1}{(n+2)(n+3)}+...\frac{1 }{(n+2)...(n+k)})<\frac{1}{(n+1)!} (1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\frac{1}{(n +1)^{k}})<\frac{1}{n!n} [/tex]

[tex]<\frac{1}{n}<\varepsilon

[/tex]

Latex is working again


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