Question #189948

(i) Show that the sequence {n/(2n-1)}_(n=1)^∞ is monotonic. (ii) Is it bounded above and/or below? Give a reason for your answer. (iii) Can you concluded that it converges? Give a reason for your answer.


1
Expert's answer
2021-05-10T15:40:35-0400

i)Given,an=n2n1.anan+1=n2n1n+12n+1.=2n2n2n2+n2n+1(2n1)(2n+1)=12n+1<0, for all n    an<an+1    anis monotonic.ii)a1=13a2=25a3=37...Since,anisincreasingsequence.Then,liman=12.Therefore, the given sequence islower bounded by 13and upper bounded by 0.5.iii)Since, the given sequence is monotonic and bounded.Therefore, it is convergent.i)\newline Given, a_n=\frac{n}{2n-1}.\newline a_n-a_{n+1}=\frac{n}{2n-1}-\frac{n+1}{2n+1}.\newline =\frac{2n^2-n-2n^2+n-2n+1}{(2n-1)(2n+1)}\newline =\frac{-1}{2n+1}<0,\space \text{for all n}\newline \implies a_n<a_{n+1}\newline \implies a_n \text{is monotonic.}\newline ii)\newline a_1=\frac{1}{3}\newline a_2=\frac{2}{5}\newline a_3=\frac{3}{7}\newline ...\newline Since, a_n is increasing sequence.\newline Then, lim a_n=\frac{1}{2}.\newline \text{Therefore, the given sequence islower bounded by }\frac{1}{3}\newline \text{and upper bounded by 0.5.}\newline iii)\newline \text{Since, the given sequence is monotonic and bounded.}\newline \text{Therefore, it is convergent.}




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