Is the sequence $f_n(x) = n(sqrt [n]x-1)$ converges uniformly on $[1,a]$ where $a >1$?

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Is the sequence $f_n(x) = n(sqrt [n]x-1)$ converges uniformly on $[1,a]$ where $a >1$ ?




My attempt: I was thinking about $M_n$ test
$$M_n = sup_x in [1,a] | n( sqrt [n]x-1)- 0|=sup_x in [1,a] |n(e^frac1nlog x -1)- 0|$$
After that i can not able to proceed further. Thanks.






real-analysis







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    up vote
    0
    down vote

    favorite













    Is the sequence $f_n(x) = n(sqrt [n]x-1)$ converges uniformly on $[1,a]$ where $a >1$ ?




    My attempt: I was thinking about $M_n$ test
    $$M_n = sup_x in [1,a] | n( sqrt [n]x-1)- 0|=sup_x in [1,a] |n(e^frac1nlog x -1)- 0|$$
    After that i can not able to proceed further. Thanks.






    real-analysis







    share|cite|improve this question

























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite












      Is the sequence $f_n(x) = n(sqrt [n]x-1)$ converges uniformly on $[1,a]$ where $a >1$ ?




      My attempt: I was thinking about $M_n$ test
      $$M_n = sup_x in [1,a] | n( sqrt [n]x-1)- 0|=sup_x in [1,a] |n(e^frac1nlog x -1)- 0|$$
      After that i can not able to proceed further. Thanks.






      real-analysis







      share|cite|improve this question
















      Is the sequence $f_n(x) = n(sqrt [n]x-1)$ converges uniformly on $[1,a]$ where $a >1$ ?




      My attempt: I was thinking about $M_n$ test
      $$M_n = sup_x in [1,a] | n( sqrt [n]x-1)- 0|=sup_x in [1,a] |n(e^frac1nlog x -1)- 0|$$
      After that i can not able to proceed further. Thanks.






      real-analysis




      real-analysis






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Sep 4 at 11:25









      Robert Z

      85.7k1055124




      85.7k1055124










      asked Sep 4 at 10:41









      stupid

      712112




      712112




















          1 Answer
          1






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          up vote
          3
          down vote



          accepted










          Hint. The pointwise limit in $[1,a]$ is $ln(x)$ (it is not $0$):
          $$lim_ntoinftyf_n(x)=lim_ntoinfty frac e^ln(x)/n-11/n=ln(x).$$
          Hence, following your approach, you should consider
          $$sup_x in [1,a] |n(e^ln(x)/n -1)- ln(x)|=
          nsup_t in [0,fracln(a)n] |e^t -1- t|.$$
          Can you take it from here?



          Alternatively, you may that for any $xin [1,a]$ the sequence $f_n(x)_ngeq 1$ is decreasing. Then use the Dini's Theorem.






          share|cite|improve this answer






















          • thanks @Robert oh yaa i misses that...but $sup_x in [1,a] |e^frac1nlog x -1)- logx|$.... how can i proceed further??
            – stupid
            Sep 4 at 11:13











          • @stupid See my edit.
            – Robert Z
            Sep 4 at 11:45










          • thanksu @Robert
            – stupid
            Sep 4 at 12:03










          Your Answer




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          active

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          3
          down vote



          accepted










          Hint. The pointwise limit in $[1,a]$ is $ln(x)$ (it is not $0$):
          $$lim_ntoinftyf_n(x)=lim_ntoinfty frac e^ln(x)/n-11/n=ln(x).$$
          Hence, following your approach, you should consider
          $$sup_x in [1,a] |n(e^ln(x)/n -1)- ln(x)|=
          nsup_t in [0,fracln(a)n] |e^t -1- t|.$$
          Can you take it from here?



          Alternatively, you may that for any $xin [1,a]$ the sequence $f_n(x)_ngeq 1$ is decreasing. Then use the Dini's Theorem.






          share|cite|improve this answer






















          • thanks @Robert oh yaa i misses that...but $sup_x in [1,a] |e^frac1nlog x -1)- logx|$.... how can i proceed further??
            – stupid
            Sep 4 at 11:13











          • @stupid See my edit.
            – Robert Z
            Sep 4 at 11:45










          • thanksu @Robert
            – stupid
            Sep 4 at 12:03














          up vote
          3
          down vote



          accepted










          Hint. The pointwise limit in $[1,a]$ is $ln(x)$ (it is not $0$):
          $$lim_ntoinftyf_n(x)=lim_ntoinfty frac e^ln(x)/n-11/n=ln(x).$$
          Hence, following your approach, you should consider
          $$sup_x in [1,a] |n(e^ln(x)/n -1)- ln(x)|=
          nsup_t in [0,fracln(a)n] |e^t -1- t|.$$
          Can you take it from here?



          Alternatively, you may that for any $xin [1,a]$ the sequence $f_n(x)_ngeq 1$ is decreasing. Then use the Dini's Theorem.






          share|cite|improve this answer






















          • thanks @Robert oh yaa i misses that...but $sup_x in [1,a] |e^frac1nlog x -1)- logx|$.... how can i proceed further??
            – stupid
            Sep 4 at 11:13











          • @stupid See my edit.
            – Robert Z
            Sep 4 at 11:45










          • thanksu @Robert
            – stupid
            Sep 4 at 12:03












          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          Hint. The pointwise limit in $[1,a]$ is $ln(x)$ (it is not $0$):
          $$lim_ntoinftyf_n(x)=lim_ntoinfty frac e^ln(x)/n-11/n=ln(x).$$
          Hence, following your approach, you should consider
          $$sup_x in [1,a] |n(e^ln(x)/n -1)- ln(x)|=
          nsup_t in [0,fracln(a)n] |e^t -1- t|.$$
          Can you take it from here?



          Alternatively, you may that for any $xin [1,a]$ the sequence $f_n(x)_ngeq 1$ is decreasing. Then use the Dini's Theorem.






          share|cite|improve this answer














          Hint. The pointwise limit in $[1,a]$ is $ln(x)$ (it is not $0$):
          $$lim_ntoinftyf_n(x)=lim_ntoinfty frac e^ln(x)/n-11/n=ln(x).$$
          Hence, following your approach, you should consider
          $$sup_x in [1,a] |n(e^ln(x)/n -1)- ln(x)|=
          nsup_t in [0,fracln(a)n] |e^t -1- t|.$$
          Can you take it from here?



          Alternatively, you may that for any $xin [1,a]$ the sequence $f_n(x)_ngeq 1$ is decreasing. Then use the Dini's Theorem.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Sep 4 at 11:45

























          answered Sep 4 at 11:03









          Robert Z

          85.7k1055124




          85.7k1055124











          • thanks @Robert oh yaa i misses that...but $sup_x in [1,a] |e^frac1nlog x -1)- logx|$.... how can i proceed further??
            – stupid
            Sep 4 at 11:13











          • @stupid See my edit.
            – Robert Z
            Sep 4 at 11:45










          • thanksu @Robert
            – stupid
            Sep 4 at 12:03
















          • thanks @Robert oh yaa i misses that...but $sup_x in [1,a] |e^frac1nlog x -1)- logx|$.... how can i proceed further??
            – stupid
            Sep 4 at 11:13











          • @stupid See my edit.
            – Robert Z
            Sep 4 at 11:45










          • thanksu @Robert
            – stupid
            Sep 4 at 12:03















          thanks @Robert oh yaa i misses that...but $sup_x in [1,a] |e^frac1nlog x -1)- logx|$.... how can i proceed further??
          – stupid
          Sep 4 at 11:13





          thanks @Robert oh yaa i misses that...but $sup_x in [1,a] |e^frac1nlog x -1)- logx|$.... how can i proceed further??
          – stupid
          Sep 4 at 11:13













          @stupid See my edit.
          – Robert Z
          Sep 4 at 11:45




          @stupid See my edit.
          – Robert Z
          Sep 4 at 11:45












          thanksu @Robert
          – stupid
          Sep 4 at 12:03




          thanksu @Robert
          – stupid
          Sep 4 at 12:03

















           

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