How to Show Sequence is Cauchy for n in the Naturals

[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.

 

[Mark44]

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?

 

[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.

 

[Mark44]

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)!

 

[boombaby]

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)!}&lt;\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

 

[Unassuming]

\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...+\frac{1}{(n+k)!}&lt;\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)!} | &lt; e

I know that can't work because there is an n in the numerator but what else can I do?

 

[Unassuming]

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]

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)})&lt;\frac{1}{(n+1)!} (1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\frac{1}{(n+1)^{k}})&lt;\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.

 

[dirk_mec1]

<br /> \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)})&lt;\frac{1}{(n+1)!} (1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\frac{1}{(n +1)^{k}})&lt; <br />
&lt; \frac{1}{(n+1)!} \sum_{j=0}^{\infty} \frac{1}{(n+1)^j} &lt; \sum_{j=0}^{\infty} \frac{1}{(n+1)^j} = \frac{n+1}{n}

 

[boombaby]

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

 

[Unassuming]

Choose N> 1/e. Then for all n > N we have that,

<br /> <br /> \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)})&lt;\frac{1}{(n+1)!} (1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\frac{1}{(n+1)^{k}})&lt;\frac{1}{n!n}&lt;\frac{1}{n}&lt;\varepsilon<br /> <br />

Therefore, |s_{n+k}-s_n|&lt;\varepsilon.

Is that right?

EDIT: Is that LaTeX showing up for anybody else?

 

[dirk_mec1]

<br /> \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)})&lt;\frac{1}{(n+1)!} (1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\frac{1}{(n +1)^{k}})&lt;\frac{1}{n!n}&lt;\frac{1}{n}&lt;\varepsilon<br />

\\Edit: indeed latex isn't working

 

[boombaby]

test:
\frac{1}{n!} \int^{1}_{0} sin x dx

Edit: indeed

 

[dirk_mec1]

<br /> <br /> \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)})&lt;\frac{1}{(n+1)!} (1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\frac{1}{(n +1)^{k}})&lt;\frac{1}{n!n}

&lt;\frac{1}{n}&lt;\varepsilon<br /> <br />

Latex is working again