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.
For the comparison route I started out doing (had not gotten to the integration of cost yet: 
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?
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, $\int_1^n \frac{cost}{t^2}dt=\sum_{t=1}^n \frac{cost}{t^2}$.