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Show sequence is cauchy 
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#1
Oct1608, 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+k1)! . I am not sure if that is worthwhile or I am totally off. 


#2
Oct1608, 09:52 PM

Mentor
P: 21,286




#3
Oct1608, 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. 


#4
Oct1608, 11:48 PM

Mentor
P: 21,286

Show sequence is cauchy
HINT: You have k terms (count 'em!) on the right, the largest of which is 1/(n + 1)! 


#5
Oct1708, 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 


#6
Oct1708, 03:48 PM

P: 169

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? 


#7
Oct1708, 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. 


#8
Oct1708, 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. 


#9
Oct1808, 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] 


#10
Oct2008, 02:57 AM

P: 130

no...it is useless that you have "........< (n+1)/n"



#11
Oct2008, 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? 


#12
Oct2008, 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 


#13
Oct2008, 12:49 PM

P: 130

test:
[tex]\frac{1}{n!} \int^{1}_{0} sin x dx[/tex] Edit: indeed 


#14
Oct2208, 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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