Verification of the approach on finding the limit of $x_n$ as $ntoinfty$ given $x_{n+1} = sqrt[k]{5x_n}$...












2















Let $x_n$ denote a sequence, $ninBbb N$:
$$
begin{cases}
x_{n+1} = sqrt[k]{5x_n}\
x_1 = sqrt[k]{5}\
kinBbb N
end{cases}
$$
Find:
$$
lim_{ntoinfty} sqrt[k]{x_n}
$$




I would like to verify my approach on finding the limit, which doesn't seem very natural and potentially find a simpler one. I've started with writing down several first terms:
$$
x_1 = sqrt[k]{5}\
x_2 = sqrt[k]{5sqrt[k]{5}}\
x_3 = sqrt[k]{5sqrt[k]{5sqrt[k]{5}}}\
dots
$$



Rewrite as:
$$
begin{align}
x_1 &= 5^{1over k}\
x_2 &= left(5cdot 5^{1over k}right)^{1over k} = 5^{frac{k+1}{k^2}}\
&cdots\
x_n &= 5^{frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}}
end{align}
$$



This is hardly readable using powers in mathjax. Rewrite:
$$
log_5 x_n = frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}
$$



Nominator is a regular geometric series:
$$
1 + k + cdots + k^{n-2} + k^{n-1} = frac{k^n - 1}{k - 1}
$$



Thus:
$$
frac{k^n - 1}{k - 1} cdot frac{1}{k^n} = frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}tag 1
$$



Since $a^x$ is continuous we may consider the limit of $(1)$:
$$
lim_{ntoinfty} left(frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}right) = frac{1}{k-1}
$$
Which implies:
$$
lim_{ntoinfty}x_n = sqrt[k-1]{5}
$$



Is it the right way to approach this problem? Could it be simplified? Thank you!






calculus limits proof-verification







share|cite|improve this question



























    2















    Let $x_n$ denote a sequence, $ninBbb N$:
    $$
    begin{cases}
    x_{n+1} = sqrt[k]{5x_n}\
    x_1 = sqrt[k]{5}\
    kinBbb N
    end{cases}
    $$
    Find:
    $$
    lim_{ntoinfty} sqrt[k]{x_n}
    $$




    I would like to verify my approach on finding the limit, which doesn't seem very natural and potentially find a simpler one. I've started with writing down several first terms:
    $$
    x_1 = sqrt[k]{5}\
    x_2 = sqrt[k]{5sqrt[k]{5}}\
    x_3 = sqrt[k]{5sqrt[k]{5sqrt[k]{5}}}\
    dots
    $$



    Rewrite as:
    $$
    begin{align}
    x_1 &= 5^{1over k}\
    x_2 &= left(5cdot 5^{1over k}right)^{1over k} = 5^{frac{k+1}{k^2}}\
    &cdots\
    x_n &= 5^{frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}}
    end{align}
    $$



    This is hardly readable using powers in mathjax. Rewrite:
    $$
    log_5 x_n = frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}
    $$



    Nominator is a regular geometric series:
    $$
    1 + k + cdots + k^{n-2} + k^{n-1} = frac{k^n - 1}{k - 1}
    $$



    Thus:
    $$
    frac{k^n - 1}{k - 1} cdot frac{1}{k^n} = frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}tag 1
    $$



    Since $a^x$ is continuous we may consider the limit of $(1)$:
    $$
    lim_{ntoinfty} left(frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}right) = frac{1}{k-1}
    $$
    Which implies:
    $$
    lim_{ntoinfty}x_n = sqrt[k-1]{5}
    $$



    Is it the right way to approach this problem? Could it be simplified? Thank you!






    calculus limits proof-verification







    share|cite|improve this question

























      2












      2








      2








      Let $x_n$ denote a sequence, $ninBbb N$:
      $$
      begin{cases}
      x_{n+1} = sqrt[k]{5x_n}\
      x_1 = sqrt[k]{5}\
      kinBbb N
      end{cases}
      $$
      Find:
      $$
      lim_{ntoinfty} sqrt[k]{x_n}
      $$




      I would like to verify my approach on finding the limit, which doesn't seem very natural and potentially find a simpler one. I've started with writing down several first terms:
      $$
      x_1 = sqrt[k]{5}\
      x_2 = sqrt[k]{5sqrt[k]{5}}\
      x_3 = sqrt[k]{5sqrt[k]{5sqrt[k]{5}}}\
      dots
      $$



      Rewrite as:
      $$
      begin{align}
      x_1 &= 5^{1over k}\
      x_2 &= left(5cdot 5^{1over k}right)^{1over k} = 5^{frac{k+1}{k^2}}\
      &cdots\
      x_n &= 5^{frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}}
      end{align}
      $$



      This is hardly readable using powers in mathjax. Rewrite:
      $$
      log_5 x_n = frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}
      $$



      Nominator is a regular geometric series:
      $$
      1 + k + cdots + k^{n-2} + k^{n-1} = frac{k^n - 1}{k - 1}
      $$



      Thus:
      $$
      frac{k^n - 1}{k - 1} cdot frac{1}{k^n} = frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}tag 1
      $$



      Since $a^x$ is continuous we may consider the limit of $(1)$:
      $$
      lim_{ntoinfty} left(frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}right) = frac{1}{k-1}
      $$
      Which implies:
      $$
      lim_{ntoinfty}x_n = sqrt[k-1]{5}
      $$



      Is it the right way to approach this problem? Could it be simplified? Thank you!






      calculus limits proof-verification







      share|cite|improve this question














      Let $x_n$ denote a sequence, $ninBbb N$:
      $$
      begin{cases}
      x_{n+1} = sqrt[k]{5x_n}\
      x_1 = sqrt[k]{5}\
      kinBbb N
      end{cases}
      $$
      Find:
      $$
      lim_{ntoinfty} sqrt[k]{x_n}
      $$




      I would like to verify my approach on finding the limit, which doesn't seem very natural and potentially find a simpler one. I've started with writing down several first terms:
      $$
      x_1 = sqrt[k]{5}\
      x_2 = sqrt[k]{5sqrt[k]{5}}\
      x_3 = sqrt[k]{5sqrt[k]{5sqrt[k]{5}}}\
      dots
      $$



      Rewrite as:
      $$
      begin{align}
      x_1 &= 5^{1over k}\
      x_2 &= left(5cdot 5^{1over k}right)^{1over k} = 5^{frac{k+1}{k^2}}\
      &cdots\
      x_n &= 5^{frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}}
      end{align}
      $$



      This is hardly readable using powers in mathjax. Rewrite:
      $$
      log_5 x_n = frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}
      $$



      Nominator is a regular geometric series:
      $$
      1 + k + cdots + k^{n-2} + k^{n-1} = frac{k^n - 1}{k - 1}
      $$



      Thus:
      $$
      frac{k^n - 1}{k - 1} cdot frac{1}{k^n} = frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}tag 1
      $$



      Since $a^x$ is continuous we may consider the limit of $(1)$:
      $$
      lim_{ntoinfty} left(frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}right) = frac{1}{k-1}
      $$
      Which implies:
      $$
      lim_{ntoinfty}x_n = sqrt[k-1]{5}
      $$



      Is it the right way to approach this problem? Could it be simplified? Thank you!






      calculus limits proof-verification




      calculus limits proof-verification






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 4 at 11:26









      romanroman

      2,02221221




      2,02221221






















          1 Answer
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          We have $x_2 = sqrt[k]{5sqrt[k]{5}} ge sqrt[k]{5} = x_1$. If we assume that $x_{n} ge x_{n-1}$, it follows
          $sqrt[k]{5x_{n}} ge sqrt[k]{5x_{n-1}}$, or $x_{n+1} ge x_n$. Induction implies that $(x_n)_n$ is increasing.



          We have $x_1 = sqrt[k]{5} le sqrt[k-1]{5}$. If we assume $x_n le sqrt[k-1]{5}$, we get $$x_{n+1} = sqrt[k]{5x_n} le sqrt[k]{5sqrt[k-1]{5}} = left(5cdot 5^{frac1{k-1}}right)^{1/k} = sqrt[k-1]{5}$$
          Induction implies that $x_n le sqrt[k-1]{5}, forall ninmathbb{N}$.



          The sequence $(x_n)_n$ is increasing and bounded from above so it converges to some $L = lim_{ntoinfty} x_n$.



          Letting $ntoinfty$ in the relation $x_{n+1} = sqrt[k]{5x_n}$ gives
          $$L = sqrt[k]{5L} implies L(L^{k-1}-5) = 0$$



          It cannot be $L = 0$ because $x_n ge x_1 =sqrt[k]{5}, forall ninmathbb{N}$. Therefore $L = sqrt[k-1]{5}$.






          share|cite|improve this answer





















          • Thank you! Induction is indeed a better suit here.
            – roman
            Jan 4 at 11:55











          Your Answer





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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          We have $x_2 = sqrt[k]{5sqrt[k]{5}} ge sqrt[k]{5} = x_1$. If we assume that $x_{n} ge x_{n-1}$, it follows
          $sqrt[k]{5x_{n}} ge sqrt[k]{5x_{n-1}}$, or $x_{n+1} ge x_n$. Induction implies that $(x_n)_n$ is increasing.



          We have $x_1 = sqrt[k]{5} le sqrt[k-1]{5}$. If we assume $x_n le sqrt[k-1]{5}$, we get $$x_{n+1} = sqrt[k]{5x_n} le sqrt[k]{5sqrt[k-1]{5}} = left(5cdot 5^{frac1{k-1}}right)^{1/k} = sqrt[k-1]{5}$$
          Induction implies that $x_n le sqrt[k-1]{5}, forall ninmathbb{N}$.



          The sequence $(x_n)_n$ is increasing and bounded from above so it converges to some $L = lim_{ntoinfty} x_n$.



          Letting $ntoinfty$ in the relation $x_{n+1} = sqrt[k]{5x_n}$ gives
          $$L = sqrt[k]{5L} implies L(L^{k-1}-5) = 0$$



          It cannot be $L = 0$ because $x_n ge x_1 =sqrt[k]{5}, forall ninmathbb{N}$. Therefore $L = sqrt[k-1]{5}$.






          share|cite|improve this answer





















          • Thank you! Induction is indeed a better suit here.
            – roman
            Jan 4 at 11:55
















          2














          We have $x_2 = sqrt[k]{5sqrt[k]{5}} ge sqrt[k]{5} = x_1$. If we assume that $x_{n} ge x_{n-1}$, it follows
          $sqrt[k]{5x_{n}} ge sqrt[k]{5x_{n-1}}$, or $x_{n+1} ge x_n$. Induction implies that $(x_n)_n$ is increasing.



          We have $x_1 = sqrt[k]{5} le sqrt[k-1]{5}$. If we assume $x_n le sqrt[k-1]{5}$, we get $$x_{n+1} = sqrt[k]{5x_n} le sqrt[k]{5sqrt[k-1]{5}} = left(5cdot 5^{frac1{k-1}}right)^{1/k} = sqrt[k-1]{5}$$
          Induction implies that $x_n le sqrt[k-1]{5}, forall ninmathbb{N}$.



          The sequence $(x_n)_n$ is increasing and bounded from above so it converges to some $L = lim_{ntoinfty} x_n$.



          Letting $ntoinfty$ in the relation $x_{n+1} = sqrt[k]{5x_n}$ gives
          $$L = sqrt[k]{5L} implies L(L^{k-1}-5) = 0$$



          It cannot be $L = 0$ because $x_n ge x_1 =sqrt[k]{5}, forall ninmathbb{N}$. Therefore $L = sqrt[k-1]{5}$.






          share|cite|improve this answer





















          • Thank you! Induction is indeed a better suit here.
            – roman
            Jan 4 at 11:55














          2












          2








          2






          We have $x_2 = sqrt[k]{5sqrt[k]{5}} ge sqrt[k]{5} = x_1$. If we assume that $x_{n} ge x_{n-1}$, it follows
          $sqrt[k]{5x_{n}} ge sqrt[k]{5x_{n-1}}$, or $x_{n+1} ge x_n$. Induction implies that $(x_n)_n$ is increasing.



          We have $x_1 = sqrt[k]{5} le sqrt[k-1]{5}$. If we assume $x_n le sqrt[k-1]{5}$, we get $$x_{n+1} = sqrt[k]{5x_n} le sqrt[k]{5sqrt[k-1]{5}} = left(5cdot 5^{frac1{k-1}}right)^{1/k} = sqrt[k-1]{5}$$
          Induction implies that $x_n le sqrt[k-1]{5}, forall ninmathbb{N}$.



          The sequence $(x_n)_n$ is increasing and bounded from above so it converges to some $L = lim_{ntoinfty} x_n$.



          Letting $ntoinfty$ in the relation $x_{n+1} = sqrt[k]{5x_n}$ gives
          $$L = sqrt[k]{5L} implies L(L^{k-1}-5) = 0$$



          It cannot be $L = 0$ because $x_n ge x_1 =sqrt[k]{5}, forall ninmathbb{N}$. Therefore $L = sqrt[k-1]{5}$.






          share|cite|improve this answer












          We have $x_2 = sqrt[k]{5sqrt[k]{5}} ge sqrt[k]{5} = x_1$. If we assume that $x_{n} ge x_{n-1}$, it follows
          $sqrt[k]{5x_{n}} ge sqrt[k]{5x_{n-1}}$, or $x_{n+1} ge x_n$. Induction implies that $(x_n)_n$ is increasing.



          We have $x_1 = sqrt[k]{5} le sqrt[k-1]{5}$. If we assume $x_n le sqrt[k-1]{5}$, we get $$x_{n+1} = sqrt[k]{5x_n} le sqrt[k]{5sqrt[k-1]{5}} = left(5cdot 5^{frac1{k-1}}right)^{1/k} = sqrt[k-1]{5}$$
          Induction implies that $x_n le sqrt[k-1]{5}, forall ninmathbb{N}$.



          The sequence $(x_n)_n$ is increasing and bounded from above so it converges to some $L = lim_{ntoinfty} x_n$.



          Letting $ntoinfty$ in the relation $x_{n+1} = sqrt[k]{5x_n}$ gives
          $$L = sqrt[k]{5L} implies L(L^{k-1}-5) = 0$$



          It cannot be $L = 0$ because $x_n ge x_1 =sqrt[k]{5}, forall ninmathbb{N}$. Therefore $L = sqrt[k-1]{5}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 11:40









          mechanodroidmechanodroid

          27k62446




          27k62446












          • Thank you! Induction is indeed a better suit here.
            – roman
            Jan 4 at 11:55


















          • Thank you! Induction is indeed a better suit here.
            – roman
            Jan 4 at 11:55
















          Thank you! Induction is indeed a better suit here.
          – roman
          Jan 4 at 11:55




          Thank you! Induction is indeed a better suit here.
          – roman
          Jan 4 at 11:55


















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