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

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

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

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

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

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

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

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