Proof verification for $lim_{ntoinfty}frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}$












2












$begingroup$



Find the limit:
$$
lim_{ntoinfty}frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} text{for} a >0
$$




Let:
$$
begin{cases}
x_n = frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}\
ninmathbb N
end{cases}
$$



By definition of exponent we may rewrite $x_n$ as:
$$
x_n = frac{a}{1+a}cdotfrac{a}{1+a^2}cdotsfrac{a}{1+a^n}
$$



Since $a>0$:
$$
forall a>0,kinmathbb N: frac{a}{1+a^k} < 1
$$



Choose $q_n$ to be in the following form:
$$
q_n = maxleft{frac{a}{1+a}, frac{a}{1+a^2}, dots,frac{a}{1+a^n}right}
$$



It follows that:
$$
x_n = frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} le q^n_n
$$
Also $x_n > 0$. Now squeezing $x_n$ one may obtain:



$$
0 le lim_{ntoinfty} x_n le lim_{ntoinfty} q^n_n = 0
$$
Thus:
$$
lim_{ntoinfty}frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} = 0
$$



Have i missed something?






calculus limits proof-verification limits-without-lhopital







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The logic is fine for me. Two notation details: when you define a sequence (like your $x_n$) don't forget the scope of $n$ ('for all $ninBbb N'$, for example). And the equation after 'Since $a>0$' doesn't need any bracket.
    $endgroup$
    – ajotatxe
    Dec 5 '18 at 15:52








  • 2




    $begingroup$
    The $q$, defined as the maximum over $frac a{1+a},...,frac a{1+a^n}$ depends on $n$, therefore it should be written as $q_n$. Then, it is not clear whether $q_n^n$ goes to zero. This is an issue.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 5 '18 at 15:53








  • 1




    $begingroup$
    One more detail: $x_nle q_n$ (strict inequality may be false, and you don't need it).
    $endgroup$
    – ajotatxe
    Dec 5 '18 at 15:54










  • $begingroup$
    Thank you for the notices
    $endgroup$
    – roman
    Dec 5 '18 at 15:57
















2












$begingroup$



Find the limit:
$$
lim_{ntoinfty}frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} text{for} a >0
$$




Let:
$$
begin{cases}
x_n = frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}\
ninmathbb N
end{cases}
$$



By definition of exponent we may rewrite $x_n$ as:
$$
x_n = frac{a}{1+a}cdotfrac{a}{1+a^2}cdotsfrac{a}{1+a^n}
$$



Since $a>0$:
$$
forall a>0,kinmathbb N: frac{a}{1+a^k} < 1
$$



Choose $q_n$ to be in the following form:
$$
q_n = maxleft{frac{a}{1+a}, frac{a}{1+a^2}, dots,frac{a}{1+a^n}right}
$$



It follows that:
$$
x_n = frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} le q^n_n
$$
Also $x_n > 0$. Now squeezing $x_n$ one may obtain:



$$
0 le lim_{ntoinfty} x_n le lim_{ntoinfty} q^n_n = 0
$$
Thus:
$$
lim_{ntoinfty}frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} = 0
$$



Have i missed something?






calculus limits proof-verification limits-without-lhopital







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The logic is fine for me. Two notation details: when you define a sequence (like your $x_n$) don't forget the scope of $n$ ('for all $ninBbb N'$, for example). And the equation after 'Since $a>0$' doesn't need any bracket.
    $endgroup$
    – ajotatxe
    Dec 5 '18 at 15:52








  • 2




    $begingroup$
    The $q$, defined as the maximum over $frac a{1+a},...,frac a{1+a^n}$ depends on $n$, therefore it should be written as $q_n$. Then, it is not clear whether $q_n^n$ goes to zero. This is an issue.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 5 '18 at 15:53








  • 1




    $begingroup$
    One more detail: $x_nle q_n$ (strict inequality may be false, and you don't need it).
    $endgroup$
    – ajotatxe
    Dec 5 '18 at 15:54










  • $begingroup$
    Thank you for the notices
    $endgroup$
    – roman
    Dec 5 '18 at 15:57














2












2








2


2



$begingroup$



Find the limit:
$$
lim_{ntoinfty}frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} text{for} a >0
$$




Let:
$$
begin{cases}
x_n = frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}\
ninmathbb N
end{cases}
$$



By definition of exponent we may rewrite $x_n$ as:
$$
x_n = frac{a}{1+a}cdotfrac{a}{1+a^2}cdotsfrac{a}{1+a^n}
$$



Since $a>0$:
$$
forall a>0,kinmathbb N: frac{a}{1+a^k} < 1
$$



Choose $q_n$ to be in the following form:
$$
q_n = maxleft{frac{a}{1+a}, frac{a}{1+a^2}, dots,frac{a}{1+a^n}right}
$$



It follows that:
$$
x_n = frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} le q^n_n
$$
Also $x_n > 0$. Now squeezing $x_n$ one may obtain:



$$
0 le lim_{ntoinfty} x_n le lim_{ntoinfty} q^n_n = 0
$$
Thus:
$$
lim_{ntoinfty}frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} = 0
$$



Have i missed something?






calculus limits proof-verification limits-without-lhopital







share|cite|improve this question











$endgroup$





Find the limit:
$$
lim_{ntoinfty}frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} text{for} a >0
$$




Let:
$$
begin{cases}
x_n = frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}\
ninmathbb N
end{cases}
$$



By definition of exponent we may rewrite $x_n$ as:
$$
x_n = frac{a}{1+a}cdotfrac{a}{1+a^2}cdotsfrac{a}{1+a^n}
$$



Since $a>0$:
$$
forall a>0,kinmathbb N: frac{a}{1+a^k} < 1
$$



Choose $q_n$ to be in the following form:
$$
q_n = maxleft{frac{a}{1+a}, frac{a}{1+a^2}, dots,frac{a}{1+a^n}right}
$$



It follows that:
$$
x_n = frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} le q^n_n
$$
Also $x_n > 0$. Now squeezing $x_n$ one may obtain:



$$
0 le lim_{ntoinfty} x_n le lim_{ntoinfty} q^n_n = 0
$$
Thus:
$$
lim_{ntoinfty}frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)} = 0
$$



Have i missed something?






calculus limits proof-verification limits-without-lhopital




calculus limits proof-verification limits-without-lhopital






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 15:57







roman

















asked Dec 5 '18 at 15:46









romanroman

2,31121224




2,31121224








  • 1




    $begingroup$
    The logic is fine for me. Two notation details: when you define a sequence (like your $x_n$) don't forget the scope of $n$ ('for all $ninBbb N'$, for example). And the equation after 'Since $a>0$' doesn't need any bracket.
    $endgroup$
    – ajotatxe
    Dec 5 '18 at 15:52








  • 2




    $begingroup$
    The $q$, defined as the maximum over $frac a{1+a},...,frac a{1+a^n}$ depends on $n$, therefore it should be written as $q_n$. Then, it is not clear whether $q_n^n$ goes to zero. This is an issue.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 5 '18 at 15:53








  • 1




    $begingroup$
    One more detail: $x_nle q_n$ (strict inequality may be false, and you don't need it).
    $endgroup$
    – ajotatxe
    Dec 5 '18 at 15:54










  • $begingroup$
    Thank you for the notices
    $endgroup$
    – roman
    Dec 5 '18 at 15:57














  • 1




    $begingroup$
    The logic is fine for me. Two notation details: when you define a sequence (like your $x_n$) don't forget the scope of $n$ ('for all $ninBbb N'$, for example). And the equation after 'Since $a>0$' doesn't need any bracket.
    $endgroup$
    – ajotatxe
    Dec 5 '18 at 15:52








  • 2




    $begingroup$
    The $q$, defined as the maximum over $frac a{1+a},...,frac a{1+a^n}$ depends on $n$, therefore it should be written as $q_n$. Then, it is not clear whether $q_n^n$ goes to zero. This is an issue.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 5 '18 at 15:53








  • 1




    $begingroup$
    One more detail: $x_nle q_n$ (strict inequality may be false, and you don't need it).
    $endgroup$
    – ajotatxe
    Dec 5 '18 at 15:54










  • $begingroup$
    Thank you for the notices
    $endgroup$
    – roman
    Dec 5 '18 at 15:57








1




1




$begingroup$
The logic is fine for me. Two notation details: when you define a sequence (like your $x_n$) don't forget the scope of $n$ ('for all $ninBbb N'$, for example). And the equation after 'Since $a>0$' doesn't need any bracket.
$endgroup$
– ajotatxe
Dec 5 '18 at 15:52






$begingroup$
The logic is fine for me. Two notation details: when you define a sequence (like your $x_n$) don't forget the scope of $n$ ('for all $ninBbb N'$, for example). And the equation after 'Since $a>0$' doesn't need any bracket.
$endgroup$
– ajotatxe
Dec 5 '18 at 15:52






2




2




$begingroup$
The $q$, defined as the maximum over $frac a{1+a},...,frac a{1+a^n}$ depends on $n$, therefore it should be written as $q_n$. Then, it is not clear whether $q_n^n$ goes to zero. This is an issue.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 15:53






$begingroup$
The $q$, defined as the maximum over $frac a{1+a},...,frac a{1+a^n}$ depends on $n$, therefore it should be written as $q_n$. Then, it is not clear whether $q_n^n$ goes to zero. This is an issue.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 15:53






1




1




$begingroup$
One more detail: $x_nle q_n$ (strict inequality may be false, and you don't need it).
$endgroup$
– ajotatxe
Dec 5 '18 at 15:54




$begingroup$
One more detail: $x_nle q_n$ (strict inequality may be false, and you don't need it).
$endgroup$
– ajotatxe
Dec 5 '18 at 15:54












$begingroup$
Thank you for the notices
$endgroup$
– roman
Dec 5 '18 at 15:57




$begingroup$
Thank you for the notices
$endgroup$
– roman
Dec 5 '18 at 15:57










2 Answers
2






active

oldest

votes


















4












$begingroup$

HINT



As an alternative by ratio test



$$frac{x_{n+1}}{x_n}=frac{a^{n+1}}{(1+a)(1+a^2)cdots(1+a^{n+1})}frac{(1+a)(1+a^2)cdots(1+a^n)}{a^n}=frac{a}{1+a^{n+1}}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I find this approach much more elegant, thanks!
    $endgroup$
    – roman
    Dec 5 '18 at 15:58










  • $begingroup$
    @roman Yes the good old ratio test seems very effective here :)
    $endgroup$
    – gimusi
    Dec 5 '18 at 15:59






  • 2




    $begingroup$
    Nice application of the ratio test (+1).
    $endgroup$
    – Robert Z
    Dec 5 '18 at 16:03










  • $begingroup$
    @RobertZ Thanks a lot much appreciative from you! Regards
    $endgroup$
    – gimusi
    Dec 5 '18 at 16:05



















2












$begingroup$

Following your idea to apply the squeeze theorem, the proof becomes simpler if we note that $0leq a<1$,
$$0leq frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}leq a^nto 0$$
and for $ageq 1$,
$$0leq frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}leq frac{a^n}{2^n a^{n(n+1)/4}}to 0$$
where we used the fact that $1+a^kgeq 2a^{k/2}.$






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    HINT



    As an alternative by ratio test



    $$frac{x_{n+1}}{x_n}=frac{a^{n+1}}{(1+a)(1+a^2)cdots(1+a^{n+1})}frac{(1+a)(1+a^2)cdots(1+a^n)}{a^n}=frac{a}{1+a^{n+1}}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I find this approach much more elegant, thanks!
      $endgroup$
      – roman
      Dec 5 '18 at 15:58










    • $begingroup$
      @roman Yes the good old ratio test seems very effective here :)
      $endgroup$
      – gimusi
      Dec 5 '18 at 15:59






    • 2




      $begingroup$
      Nice application of the ratio test (+1).
      $endgroup$
      – Robert Z
      Dec 5 '18 at 16:03










    • $begingroup$
      @RobertZ Thanks a lot much appreciative from you! Regards
      $endgroup$
      – gimusi
      Dec 5 '18 at 16:05
















    4












    $begingroup$

    HINT



    As an alternative by ratio test



    $$frac{x_{n+1}}{x_n}=frac{a^{n+1}}{(1+a)(1+a^2)cdots(1+a^{n+1})}frac{(1+a)(1+a^2)cdots(1+a^n)}{a^n}=frac{a}{1+a^{n+1}}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I find this approach much more elegant, thanks!
      $endgroup$
      – roman
      Dec 5 '18 at 15:58










    • $begingroup$
      @roman Yes the good old ratio test seems very effective here :)
      $endgroup$
      – gimusi
      Dec 5 '18 at 15:59






    • 2




      $begingroup$
      Nice application of the ratio test (+1).
      $endgroup$
      – Robert Z
      Dec 5 '18 at 16:03










    • $begingroup$
      @RobertZ Thanks a lot much appreciative from you! Regards
      $endgroup$
      – gimusi
      Dec 5 '18 at 16:05














    4












    4








    4





    $begingroup$

    HINT



    As an alternative by ratio test



    $$frac{x_{n+1}}{x_n}=frac{a^{n+1}}{(1+a)(1+a^2)cdots(1+a^{n+1})}frac{(1+a)(1+a^2)cdots(1+a^n)}{a^n}=frac{a}{1+a^{n+1}}$$






    share|cite|improve this answer









    $endgroup$



    HINT



    As an alternative by ratio test



    $$frac{x_{n+1}}{x_n}=frac{a^{n+1}}{(1+a)(1+a^2)cdots(1+a^{n+1})}frac{(1+a)(1+a^2)cdots(1+a^n)}{a^n}=frac{a}{1+a^{n+1}}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 5 '18 at 15:57









    gimusigimusi

    92.9k84494




    92.9k84494












    • $begingroup$
      I find this approach much more elegant, thanks!
      $endgroup$
      – roman
      Dec 5 '18 at 15:58










    • $begingroup$
      @roman Yes the good old ratio test seems very effective here :)
      $endgroup$
      – gimusi
      Dec 5 '18 at 15:59






    • 2




      $begingroup$
      Nice application of the ratio test (+1).
      $endgroup$
      – Robert Z
      Dec 5 '18 at 16:03










    • $begingroup$
      @RobertZ Thanks a lot much appreciative from you! Regards
      $endgroup$
      – gimusi
      Dec 5 '18 at 16:05


















    • $begingroup$
      I find this approach much more elegant, thanks!
      $endgroup$
      – roman
      Dec 5 '18 at 15:58










    • $begingroup$
      @roman Yes the good old ratio test seems very effective here :)
      $endgroup$
      – gimusi
      Dec 5 '18 at 15:59






    • 2




      $begingroup$
      Nice application of the ratio test (+1).
      $endgroup$
      – Robert Z
      Dec 5 '18 at 16:03










    • $begingroup$
      @RobertZ Thanks a lot much appreciative from you! Regards
      $endgroup$
      – gimusi
      Dec 5 '18 at 16:05
















    $begingroup$
    I find this approach much more elegant, thanks!
    $endgroup$
    – roman
    Dec 5 '18 at 15:58




    $begingroup$
    I find this approach much more elegant, thanks!
    $endgroup$
    – roman
    Dec 5 '18 at 15:58












    $begingroup$
    @roman Yes the good old ratio test seems very effective here :)
    $endgroup$
    – gimusi
    Dec 5 '18 at 15:59




    $begingroup$
    @roman Yes the good old ratio test seems very effective here :)
    $endgroup$
    – gimusi
    Dec 5 '18 at 15:59




    2




    2




    $begingroup$
    Nice application of the ratio test (+1).
    $endgroup$
    – Robert Z
    Dec 5 '18 at 16:03




    $begingroup$
    Nice application of the ratio test (+1).
    $endgroup$
    – Robert Z
    Dec 5 '18 at 16:03












    $begingroup$
    @RobertZ Thanks a lot much appreciative from you! Regards
    $endgroup$
    – gimusi
    Dec 5 '18 at 16:05




    $begingroup$
    @RobertZ Thanks a lot much appreciative from you! Regards
    $endgroup$
    – gimusi
    Dec 5 '18 at 16:05











    2












    $begingroup$

    Following your idea to apply the squeeze theorem, the proof becomes simpler if we note that $0leq a<1$,
    $$0leq frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}leq a^nto 0$$
    and for $ageq 1$,
    $$0leq frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}leq frac{a^n}{2^n a^{n(n+1)/4}}to 0$$
    where we used the fact that $1+a^kgeq 2a^{k/2}.$






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Following your idea to apply the squeeze theorem, the proof becomes simpler if we note that $0leq a<1$,
      $$0leq frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}leq a^nto 0$$
      and for $ageq 1$,
      $$0leq frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}leq frac{a^n}{2^n a^{n(n+1)/4}}to 0$$
      where we used the fact that $1+a^kgeq 2a^{k/2}.$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Following your idea to apply the squeeze theorem, the proof becomes simpler if we note that $0leq a<1$,
        $$0leq frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}leq a^nto 0$$
        and for $ageq 1$,
        $$0leq frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}leq frac{a^n}{2^n a^{n(n+1)/4}}to 0$$
        where we used the fact that $1+a^kgeq 2a^{k/2}.$






        share|cite|improve this answer











        $endgroup$



        Following your idea to apply the squeeze theorem, the proof becomes simpler if we note that $0leq a<1$,
        $$0leq frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}leq a^nto 0$$
        and for $ageq 1$,
        $$0leq frac{a^n}{(1+a)(1+a^2)cdots(1+a^n)}leq frac{a^n}{2^n a^{n(n+1)/4}}to 0$$
        where we used the fact that $1+a^kgeq 2a^{k/2}.$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 5 '18 at 16:05

























        answered Dec 5 '18 at 16:00









        Robert ZRobert Z

        99.1k1068139




        99.1k1068139






























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