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I am looking at this problem and making sure I did it correctly. (Note I believe I need to add an n+1 at the end of the sum so I plan on doing that). I have another idea to solve this problem which answers the question using the Comparison Theorem. This route is much quicker. I know Cauchy means we need to show it converges.
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For the comparison route I started out doing (had not gotten to the integration of cost yet: enter image description here

So I am asking is my Riemann sum series valid option to show that it's Cauchy? Or should I stick with the comparison route?

Natasha Epperson
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By definition of Cauchy sequence, we need to show $\forall \epsilon>0$, $\exists N>0$, such that $\forall m,n>N,|x_n-x_m|<\epsilon$.

Note for $m>n\ge N$,

$|x_n-x_m|=|\int_n^m \frac{cost}{t^2}dt|\le \int_n^m|\frac{cost}{t^2}|dt\le \int_n^m|\frac{1}{t^2}|dt=\frac{1}{n}-\frac{1}{m}=\frac{m-n}{mn}\le \frac{1}{n}\le \frac{1}{N}$

Hence for large enough $N$, for any $m,n>N$, we have $|x_n-x_m|<\epsilon$, which shows the sequence is Cauchy.

This method's idea is similar to your comparison method. For your Riemann sum method, I am not sure how do you get the second identity, $\int_1^n \frac{cost}{t^2}dt=\sum_{t=1}^n \frac{cost}{t^2}$.

John
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  • Ohh! I was looking at doing it that way too but when I was looking at the Riemann's Integrals I kept thinking I needed to solve the integral and got stuck. Cause I know which Cauchy you need to show the inequalities. – Natasha Epperson May 03 '15 at 14:09
  • For the RSM that was a total guess heh. That's why I asked if that method was possible – Natasha Epperson May 03 '15 at 14:11
  • @NatashaEpperson "Cause I know which Cauchy you need to show the inequalities." I don't quite understand your meaning. – John May 03 '15 at 14:21
  • What i mean is I was showing the inequalities but I thought I had to solve the integrals with the inequalities. So I thought i need to do |costt/t|...<|whatever the solution is| and than separate the inequalities there. Cause I know when dealing with triangle inequalities like say a quadratic you do this: |x_n-x_m|=|ax^+bx+c|=.... – Natasha Epperson May 03 '15 at 15:06