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 September 4th, 2019, 03:24 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 05:17 PM. September 6th, 2019, 08:20 PM #2 Global Moderator   Joined: May 2007 Posts: 6,852 Thanks: 743 Hint $b_n$ series converges. $c_n$ series diverges. Tags cauchy, criterion, series Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post gusrianputra Real Analysis 4 October 25th, 2018 11:09 PM MaxBenjamin Probability and Statistics 0 July 30th, 2017 02:33 PM Doctorfunk_15 Advanced Statistics 1 June 23rd, 2012 03:20 PM pedja Number Theory 0 February 22nd, 2012 06:16 AM xsw001 Real Analysis 0 January 23rd, 2011 04:28 PM

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