0 votes 0 votes For $n\geq 1$, let $a_n=\frac{1}{2^2} + \frac{2}{3^2}+ \dots +\frac{n}{(n+1)^2}$ and $b_n=c_0 + c_1r + c_2r^2 + \dots + c_nr^n$,where $\mid c_k \mid \leq M$ for all integer $k$ and $\mid r \mid <1$. Then both $\{a_n\}$ and $\{b_n\}$ are Cauchy sequences $\{a_n\}$ is a Cauchy sequence,and $\{b_n\}$ is not Cauchy sequence $\{a_n\}$ is not a Cauchy sequence,and $\{b_n\}$ is Cauchy sequence neither $\{a_n\}$ nor $\{b_n\}$ is a Cauchy sequence. Others isi2018-mma sequence-series non-gate + – akash.dinkar12 asked May 11, 2019 • edited Nov 11, 2019 by go_editor akash.dinkar12 672 views answer comment Share Follow See 1 comment See all 1 1 comment reply NastyBall commented Jul 9, 2021 reply Follow Share One can easily show that $a_n$ is monotonically increasing unbounded sequence. Hence, it is not Cauchy. Further, one can show that $b_n$ is a bounded sequence for all $n$. Since it is given that $|c_k|\le M$ for all $k$, one can easily produce at least two limit points of the sequence $b_n$. Hence, $b_n$ is not Cauchy. Thus, option D is the correct answer. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes We use by converting it into series series n/(n+1)^2 then we see it is div. So partial sum sequence also so a_n is not cauchy but b_n is cauchy.just idea. rsquare answered Apr 4, 2022 rsquare comment Share Follow See all 0 reply Please log in or register to add a comment.