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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

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  1. graydarl Group Title
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    • one year ago
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  2. barrycarter Group Title
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    Hint: how many terms are in each element of the sequence, and what are they bounded by?

    • one year ago
  3. graydarl Group Title
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    each one of them has lim=0?

    • one year ago
  4. barrycarter Group Title
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    Not quite. There are n terms, each of which is very near 1/n

    • one year ago
  5. graydarl Group Title
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    it s true, i can see that now, but how can i use this thing?

    • one year ago
  6. zzr0ck3r Group Title
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    let x_n = 1+1/n

    • one year ago
  7. graydarl Group Title
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    x_n is just 1+1/n?

    • one year ago
  8. graydarl Group Title
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    Please explain further

    • one year ago
  9. ash2326 Group Title
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    @graydarl Have you studied sandwich theorem ?

    • one year ago
  10. ash2326 Group Title
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    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
  11. ash2326 Group Title
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    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
  12. graydarl Group Title
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    Thank you very much, i understood now, thank you

    • one year ago
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