sequence of $a^n$












0














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







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  • 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.
    – Stockfish
    Nov 29 '18 at 13:47


















0














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
























  • 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.
    – Stockfish
    Nov 29 '18 at 13:47
















0












0








0







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















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

3,08121038




3,08121038










asked Nov 29 '18 at 13:45









adsy9adsy9

32




32












  • 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.
    – Stockfish
    Nov 29 '18 at 13:47




















  • 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.
    – Stockfish
    Nov 29 '18 at 13:47


















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.
– Stockfish
Nov 29 '18 at 13:47






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.
– Stockfish
Nov 29 '18 at 13:47












2 Answers
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oldest

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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





























    0














    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





















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      2 Answers
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      2 Answers
      2






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      2














      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


























        2














        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
























          2












          2








          2






          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












          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

          14.5k3937




          14.5k3937























              0














              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


























                0














                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
























                  0












                  0








                  0






                  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












                  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

                  44.4k1845




                  44.4k1845






























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