September 4th, 2019, 02:24 PM  #1 
Member Joined: Aug 2016 From: Romania Posts: 32 Thanks: 1  Cauchy series criterion
I have two exercises: $$b_n=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+.. .+\frac{1}{n^2}$$ $$n\geq1$$ $$c_n=\frac{1}{\sqrt1}+\frac{1}{\sqrt2}+...+\frac{ 1}{\sqrt{n}}$$ $$n\geq1$$ Do I replace the terms $x_{n+p}$ and $x_n$ with: $\displaystyle \left \frac{1}{(n+p)^2}\frac{1}{n^2} \right \lt ε$ $\displaystyle \left \frac{1}{\sqrt{n+p}}\frac{1}{\sqrt {n}} \right \lt ε$ in the criterion:$\left x_{n+p}x_n \right \lt ε$? That is all I have to prove if the exercise wants me to use the convergence criterion of Cauchy? Would it be a mistake if I wrote both series like these? $\displaystyle b_{n+p} = \frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+...+\frac{1}{( n+p)^2}$ $\displaystyle c_{n+p} = \frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}+...+ \frac{1}{\sqrt{n+p}}$ Last edited by skipjack; September 4th, 2019 at 04:17 PM. 
September 6th, 2019, 07:20 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,823 Thanks: 723 
Hint $b_n$ series converges. $c_n$ series diverges.


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cauchy, criterion, series 
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