# Show sequence is cauchy

by Unassuming
Tags: cauchy, sequence
 P: 169 For n in the naturals, let $$p_n = 1 + \frac{1}{2!} + ... + \frac{1}{n!}$$ Show it is cauchy. Attempt: I have set up |p_n+k - p_n | < e , and I have solved for this. I got $$|p_{n+k} - p_n | = \frac{1}{(n+1)!} + ... + \frac{1}{(n+k)!}$$ 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.
Mentor
P: 19,802
 Quote by Unassuming For n in the naturals, let $$p_n = 1 + \frac{1}{2!} + ... + \frac{1}{n!}$$ Show it is cauchy. Attempt: I have set up |p_n+k - p_n | < e , and I have solved for this. I got $$|p_{n+k} - p_n | = \frac{1}{(n+1)!} + ... + \frac{1}{(n+k)!}$$ 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?
 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 $$\frac{1}{n!(n+k)}$$ I am doing this in order to get the telescoping sequence.
Mentor
P: 19,802

## Show sequence is cauchy

 Quote by Unassuming 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 $$\frac{1}{n!(n+k)}$$ 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)!
 P: 130 I'm not sure how to use your hint, Mark44.... are you going to prove it by saying that $$\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...+\frac{1}{(n+k)!}<\frac{k}{(n+1)!}$$, 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
P: 169
 Quote by boombaby $$\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...+\frac{1}{(n+k)!}<\frac{k}{(n+1)!}$$,
I got $$\frac{k}{(n+1)!} = \frac{k}{n!} - \frac{kn}{(n+1)!}$$

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

$$| \frac{k}{n!} - \frac{kn}{(n+1)!} | < e$$

I know that can't work because there is an n in the numerator but what else can I do?
 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.
 P: 130 this is not right.. as I said before, your k should be independent on epsilon. try $$\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...+\frac{1}{(n+k)!}=\frac{1}{(n+1)!} (1+\frac{1}{(n+2)}+\frac{1}{(n+2)(n+3)}+...\frac{1}{(n+2)...(n+k)})<\fr ac{1}{(n+1)!} (1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\frac{1}{(n+1)^{k}})<\frac{1}{n! n}$$ so we can choose a N large enough, no matter how large your k is, its sum is less than a given epislon.
 P: 623 $$\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...+\frac{1}{(n+ k)!}=\frac{1}{(n+1)!} (1+\frac{1}{(n+2)}+\frac{1}{(n+2)(n+3)}+...\frac{1 }{(n+2)...(n+k)})<\frac{1}{(n+1)!} (1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\frac{1}{(n +1)^{k}})<$$ $$< \frac{1}{(n+1)!} \sum_{j=0}^{\infty} \frac{1}{(n+1)^j} < \sum_{j=0}^{\infty} \frac{1}{(n+1)^j} = \frac{n+1}{n}$$
 P: 130 no...it is useless that you have "........< (n+1)/n"
 P: 169 Choose N> 1/e. Then for all n > N we have that, $$\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...+\frac{1}{(n+k)!}=\frac{1}{(n+1)!} (1+\frac{1}{(n+2)}+\frac{1}{(n+2)(n+3)}+...\frac{1}{(n+2)...(n+k)})<\fr ac{1}{(n+1)!} (1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\frac{1}{(n+1)^{k}})<\frac{1}{n! n}<\frac{1}{n}<\varepsilon$$ Therefore, $$|s_{n+k}-s_n|<\varepsilon$$. Is that right? EDIT: Is that LaTeX showing up for anybody else?
 P: 623 $$\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...+\frac{1}{(n+ k)!}=\frac{1}{(n+1)!} (1+\frac{1}{(n+2)}+\frac{1}{(n+2)(n+3)}+...\frac{1 }{(n+2)...(n+k)})<\frac{1}{(n+1)!} (1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\frac{1}{(n +1)^{k}})<\frac{1}{n!n}<\frac{1}{n}<\varepsilon$$ \\Edit: indeed latex isn't working
 P: 130 test: $$\frac{1}{n!} \int^{1}_{0} sin x dx$$ Edit: indeed
 P: 623 $$\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...+\frac{1}{(n+ k)!}=\frac{1}{(n+1)!} (1+\frac{1}{(n+2)}+\frac{1}{(n+2)(n+3)}+...\frac{1 }{(n+2)...(n+k)})<\frac{1}{(n+1)!} (1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\frac{1}{(n +1)^{k}})<\frac{1}{n!n}$$ $$<\frac{1}{n}<\varepsilon$$ Latex is working again

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