Worked out examples using Cauchy criterion for series

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The discussion focuses on the application of the Cauchy Criterion for Series, specifically through worked examples. Participants seek resources and demonstrate the criterion using the series \(\sum^{\infty}_{n=1} \frac{1}{n}\) and \(\sum^{\infty}_{n=1} \frac{1}{n(n+1)}\). A key method discussed involves grouping terms into \(2^n\) groups to analyze convergence, with references to Steven Lay's textbook and Wikipedia for foundational understanding. The conversation emphasizes the importance of finding suitable values for \(m\) and \(n\) to ensure that \(s_m - s_n\) remains greater than epsilon.

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michonamona
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Hello everyone,

Can anybody suggest a website that has worked out examples using the Cauchy Criterion for Series? or, if your feeling ambitious, work out the following problems below:

1.
\sum^{\infty}_{n=1}1/n

2.

\sum^{\infty}_{n=1}1/(n(n+1))The reason why I'm asking for this is because our textbook introduces the theorem (Steven Lay) but it does not demonstrate its usage. I have also perused the web and was not able to find a complete demonstration.

I only need to see it being used. The problems above are not homework questions. I promise.

I appreciate your help

M
 
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Check here for a review of the test and for reference to the variables I am referring to; http://en.wikipedia.org/wiki/Cauchy's_convergence_test.

Basically the trick is finding suitable values for m and n that make sure s_m - s_n can not be smaller than epsilon. This can be done in many ways but there's no systematic method. One way is to try to compare it to another series. The first example is a well known proof.

1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + ...+ 1/16) + (1/17+...+ 1/32) ... continually grouping them into groups of 2^n terms, you might want to investigate what each group must be larger than.

In terms of Cauchy, you are investigating s_{2^n} - s_{2^{n-1}}.
 
Thank you for your reply.

What do you mean by grouping them into 2^n groups?
 
michonamona said:
Thank you for your reply.

What do you mean by grouping them into 2^n groups?

Think he means groups of (1/2)^n terms

ie

1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + ...+ 1/16) + (1/17+...+ 1/32) ...

>

1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16 + ...+ 1/16) + (1/32+...+ 1/32) ...

And note that s2n-1 = n+1 for the second sequence of partial sums

Then think about convergence tests you know
 
Being a fool I made it harder than it had to be! Simpler case to look at is m=2n.
 

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