My Math Forum Cauchy series criterion

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 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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