sequence of $a^n$












0












$begingroup$


I am trying to show that when a $in Bbb R$ with $lvert arvert < 1$ the limit $lim_{nto+infty} a^n = 0$ using the definition of convergence of a sequence. So far I have



$lvert a^n - 0rvert < epsilon$



$lvert a^n rvert < epsilon$



$lvert arvert^n < epsilon$



and



$lvert a rvert < 1 $ $implies$ $lvert arvert^n < 1$



But I don't see where I am supposed to head with this in order to prove it? any advice would be greatly appreciated in understanding this. Thank you.






sequences-and-series analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    For given $varepsilon >0$, you have to find $N$ such that for every $n>N$ $|a|^n < varepsilon$ holds. Note that $N$ may depend on $a$. Note further that $n mapsto |a|^n$ is decreasing.
    $endgroup$
    – Stockfish
    Nov 29 '18 at 13:47


















0












$begingroup$


I am trying to show that when a $in Bbb R$ with $lvert arvert < 1$ the limit $lim_{nto+infty} a^n = 0$ using the definition of convergence of a sequence. So far I have



$lvert a^n - 0rvert < epsilon$



$lvert a^n rvert < epsilon$



$lvert arvert^n < epsilon$



and



$lvert a rvert < 1 $ $implies$ $lvert arvert^n < 1$



But I don't see where I am supposed to head with this in order to prove it? any advice would be greatly appreciated in understanding this. Thank you.






sequences-and-series analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    For given $varepsilon >0$, you have to find $N$ such that for every $n>N$ $|a|^n < varepsilon$ holds. Note that $N$ may depend on $a$. Note further that $n mapsto |a|^n$ is decreasing.
    $endgroup$
    – Stockfish
    Nov 29 '18 at 13:47
















0












0








0





$begingroup$


I am trying to show that when a $in Bbb R$ with $lvert arvert < 1$ the limit $lim_{nto+infty} a^n = 0$ using the definition of convergence of a sequence. So far I have



$lvert a^n - 0rvert < epsilon$



$lvert a^n rvert < epsilon$



$lvert arvert^n < epsilon$



and



$lvert a rvert < 1 $ $implies$ $lvert arvert^n < 1$



But I don't see where I am supposed to head with this in order to prove it? any advice would be greatly appreciated in understanding this. Thank you.






sequences-and-series analysis







share|cite|improve this question











$endgroup$




I am trying to show that when a $in Bbb R$ with $lvert arvert < 1$ the limit $lim_{nto+infty} a^n = 0$ using the definition of convergence of a sequence. So far I have



$lvert a^n - 0rvert < epsilon$



$lvert a^n rvert < epsilon$



$lvert arvert^n < epsilon$



and



$lvert a rvert < 1 $ $implies$ $lvert arvert^n < 1$



But I don't see where I am supposed to head with this in order to prove it? any advice would be greatly appreciated in understanding this. Thank you.






sequences-and-series analysis




sequences-and-series analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 14:10









Tianlalu

3,08621038




3,08621038










asked Nov 29 '18 at 13:45









adsy9adsy9

32




32












  • $begingroup$
    For given $varepsilon >0$, you have to find $N$ such that for every $n>N$ $|a|^n < varepsilon$ holds. Note that $N$ may depend on $a$. Note further that $n mapsto |a|^n$ is decreasing.
    $endgroup$
    – Stockfish
    Nov 29 '18 at 13:47




















  • $begingroup$
    For given $varepsilon >0$, you have to find $N$ such that for every $n>N$ $|a|^n < varepsilon$ holds. Note that $N$ may depend on $a$. Note further that $n mapsto |a|^n$ is decreasing.
    $endgroup$
    – Stockfish
    Nov 29 '18 at 13:47


















$begingroup$
For given $varepsilon >0$, you have to find $N$ such that for every $n>N$ $|a|^n < varepsilon$ holds. Note that $N$ may depend on $a$. Note further that $n mapsto |a|^n$ is decreasing.
$endgroup$
– Stockfish
Nov 29 '18 at 13:47






$begingroup$
For given $varepsilon >0$, you have to find $N$ such that for every $n>N$ $|a|^n < varepsilon$ holds. Note that $N$ may depend on $a$. Note further that $n mapsto |a|^n$ is decreasing.
$endgroup$
– Stockfish
Nov 29 '18 at 13:47












2 Answers
2






active

oldest

votes


















2












$begingroup$

By taking $ln$ from both sides we have $$nln |a|<ln epsilon$$and since $ln |a|<0$ (because $0<|a|<1$) we obtain $$n>{ln epsilonover ln |a|}$$therefore choose $N={ln epsilonover ln |a|}$ and you can conclude for any $n>N$ that $|a|^n<epsilon$ and the proof is complete.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    If $|a|<1$, then $1/|a|>1$, hence $frac{1}{|a|}=1+h$ with some $h>0$. Then we get, with Bernoulli's inequality:



    $frac{1}{|a^n|}=(1+h)^n ge 1+nh > nh.$



    Thus



    $|a^n| < frac{1}{nh}$.



    Can you proceed ?






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

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      2












      $begingroup$

      By taking $ln$ from both sides we have $$nln |a|<ln epsilon$$and since $ln |a|<0$ (because $0<|a|<1$) we obtain $$n>{ln epsilonover ln |a|}$$therefore choose $N={ln epsilonover ln |a|}$ and you can conclude for any $n>N$ that $|a|^n<epsilon$ and the proof is complete.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        By taking $ln$ from both sides we have $$nln |a|<ln epsilon$$and since $ln |a|<0$ (because $0<|a|<1$) we obtain $$n>{ln epsilonover ln |a|}$$therefore choose $N={ln epsilonover ln |a|}$ and you can conclude for any $n>N$ that $|a|^n<epsilon$ and the proof is complete.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          By taking $ln$ from both sides we have $$nln |a|<ln epsilon$$and since $ln |a|<0$ (because $0<|a|<1$) we obtain $$n>{ln epsilonover ln |a|}$$therefore choose $N={ln epsilonover ln |a|}$ and you can conclude for any $n>N$ that $|a|^n<epsilon$ and the proof is complete.






          share|cite|improve this answer









          $endgroup$



          By taking $ln$ from both sides we have $$nln |a|<ln epsilon$$and since $ln |a|<0$ (because $0<|a|<1$) we obtain $$n>{ln epsilonover ln |a|}$$therefore choose $N={ln epsilonover ln |a|}$ and you can conclude for any $n>N$ that $|a|^n<epsilon$ and the proof is complete.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '18 at 13:49









          Mostafa AyazMostafa Ayaz

          15.5k3939




          15.5k3939























              0












              $begingroup$

              If $|a|<1$, then $1/|a|>1$, hence $frac{1}{|a|}=1+h$ with some $h>0$. Then we get, with Bernoulli's inequality:



              $frac{1}{|a^n|}=(1+h)^n ge 1+nh > nh.$



              Thus



              $|a^n| < frac{1}{nh}$.



              Can you proceed ?






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                If $|a|<1$, then $1/|a|>1$, hence $frac{1}{|a|}=1+h$ with some $h>0$. Then we get, with Bernoulli's inequality:



                $frac{1}{|a^n|}=(1+h)^n ge 1+nh > nh.$



                Thus



                $|a^n| < frac{1}{nh}$.



                Can you proceed ?






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  If $|a|<1$, then $1/|a|>1$, hence $frac{1}{|a|}=1+h$ with some $h>0$. Then we get, with Bernoulli's inequality:



                  $frac{1}{|a^n|}=(1+h)^n ge 1+nh > nh.$



                  Thus



                  $|a^n| < frac{1}{nh}$.



                  Can you proceed ?






                  share|cite|improve this answer









                  $endgroup$



                  If $|a|<1$, then $1/|a|>1$, hence $frac{1}{|a|}=1+h$ with some $h>0$. Then we get, with Bernoulli's inequality:



                  $frac{1}{|a^n|}=(1+h)^n ge 1+nh > nh.$



                  Thus



                  $|a^n| < frac{1}{nh}$.



                  Can you proceed ?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 29 '18 at 13:51









                  FredFred

                  45.7k1848




                  45.7k1848






























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