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graydarl
Group Title
I have a sequence that must have lim when n goes to infinity equal to 1 (pic below)
 one year ago
 one year ago
graydarl Group Title
I have a sequence that must have lim when n goes to infinity equal to 1 (pic below)
 one year ago
 one year ago

This Question is Closed

barrycarter Group TitleBest ResponseYou've already chosen the best response.0
Hint: how many terms are in each element of the sequence, and what are they bounded by?
 one year ago

graydarl Group TitleBest ResponseYou've already chosen the best response.0
each one of them has lim=0?
 one year ago

barrycarter Group TitleBest ResponseYou've already chosen the best response.0
Not quite. There are n terms, each of which is very near 1/n
 one year ago

graydarl Group TitleBest ResponseYou've already chosen the best response.0
it s true, i can see that now, but how can i use this thing?
 one year ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.0
let x_n = 1+1/n
 one year ago

graydarl Group TitleBest ResponseYou've already chosen the best response.0
x_n is just 1+1/n?
 one year ago

graydarl Group TitleBest ResponseYou've already chosen the best response.0
Please explain further
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.1
@graydarl Have you studied sandwich theorem ?
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.1
We have the series as \[\lim_{n\to \infty} (\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+....\frac{1}{\sqrt{n^2+n}})\] Obviosuly each of the term is less than 1/n \[\frac{1}{\sqrt{n^2+1}}< \frac 1 n \] \[\frac{1}{\sqrt{n^2+n}}< \frac 1 n\] so whole series sum will be less than \[\frac 1 n + \frac 1 n ....\frac 1 n\] note that there are n terms And if you notice each of the term is greater than 1/(n+1) \[\frac{1}{\sqrt{n^2+1}}>\frac{1}{\sqrt{n^2+2n+1}}\] \[\frac{1}{\sqrt{n^2+n}}>\frac{1}{\sqrt{n^2+2n+1}}\] so we have \[\frac{1}{n+1}+\frac{1}{n+1}....+\frac{1}{n+1}<\frac 1 {\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}....+\frac{1}{\sqrt{n^2+n}}\] \[<\frac{1}{n}+\frac{1}{n}.........+\frac{1}{n}\] Now apply the limit \(n\to \infty\) \[\lim_{n\to \infty}(\frac{1}{n+1}+\frac{1}{n+1}....+\frac{1}{n+1})<\] \[\lim_{n\to \infty}(\frac 1 {\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}....+\frac{1}{\sqrt{n^2+n}})\] \[<\lim_{n\to \infty}(\frac{1}{n}+\frac{1}{n}.........+\frac{1}{n})\] Evaluate the limits for lower and upper bound. If you get the same limit, then by sandwich theorem, you'll be able to prove. Could you try @graydarl
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.1
I mean evaluate these two limits \[\Large {\lim_{n\to \infty}(\frac{1}{n+1}+\frac{1}{n+1}....+\frac{1}{n+1})}\] \[\Large \lim_{n\to \infty}(\frac{1}{n}+\frac{1}{n}.........+\frac{1}{n})\]
 one year ago

graydarl Group TitleBest ResponseYou've already chosen the best response.0
Thank you very much, i understood now, thank you
 one year ago
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