Closed form for sum












0












$begingroup$


I wonder whether there is a closed form for this sum
$$ S_n:=displaystylesum_{k=0}^n dfrac{4^k}{4^k+5^k}$$
The question asks to express the sum in terms of $n$ then to deduce the limit of $dfrac{S_n}{n+1}$.
I tried to use the following sum as an auxiliary sum
$$T_n:=displaystylesum_{k=0}^ndfrac{5^k}{4^k+5^k}$$
noticing that $S_n + T_n = n+1$.
Any thoughts about this ? thanks.






sequences-and-series summation telescopic-series







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    I don't know how to express the sum in terms of $n$, but I am a little confused. Don't we have $lim_{ntoinfty}S_n=sum_{k=0}^inftyfrac{4^k}{4^k+5^k}leqsum_{k=0}^infty(frac45)^k=frac1{1-frac45}=5$ and hence $lim_{ntoinfty}frac{S_n}{n+1}=0$?
    $endgroup$
    – SmileyCraft
    Dec 16 '18 at 12:49






  • 1




    $begingroup$
    Also, Wolfram Alpha gives a closed form involving the $q$-digamma function, which I've never heared of, and I doubt you have. What makes you think there is a closed form of the sum?
    $endgroup$
    – SmileyCraft
    Dec 16 '18 at 12:54










  • $begingroup$
    Thanks, the first part of the question in a paper (which I doubt is wrong) asks to express $S_n$ in terms of $n$ before computing the limit of the quotient $dfrac{S_n}{n+1}$
    $endgroup$
    – Oussama Sarih
    Dec 16 '18 at 13:00








  • 2




    $begingroup$
    @OussamaSarih, can you link to the paper, or reproduce verbatim what it says about expressing $S_n$? Given that the first three $S_n$'s are $1/2$, $17/18$, and $985/738$, I'd be a little surprised to see any kind of simple closed formula. (Even well written papers make occasional mistakes.)
    $endgroup$
    – Barry Cipra
    Dec 16 '18 at 13:04












  • $begingroup$
    It's a homework sheet in french for highschool, I'll post a screenshot as soon as I get from one from the students who asked me to solve it.
    $endgroup$
    – Oussama Sarih
    Dec 16 '18 at 15:39
















0












$begingroup$


I wonder whether there is a closed form for this sum
$$ S_n:=displaystylesum_{k=0}^n dfrac{4^k}{4^k+5^k}$$
The question asks to express the sum in terms of $n$ then to deduce the limit of $dfrac{S_n}{n+1}$.
I tried to use the following sum as an auxiliary sum
$$T_n:=displaystylesum_{k=0}^ndfrac{5^k}{4^k+5^k}$$
noticing that $S_n + T_n = n+1$.
Any thoughts about this ? thanks.






sequences-and-series summation telescopic-series







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    I don't know how to express the sum in terms of $n$, but I am a little confused. Don't we have $lim_{ntoinfty}S_n=sum_{k=0}^inftyfrac{4^k}{4^k+5^k}leqsum_{k=0}^infty(frac45)^k=frac1{1-frac45}=5$ and hence $lim_{ntoinfty}frac{S_n}{n+1}=0$?
    $endgroup$
    – SmileyCraft
    Dec 16 '18 at 12:49






  • 1




    $begingroup$
    Also, Wolfram Alpha gives a closed form involving the $q$-digamma function, which I've never heared of, and I doubt you have. What makes you think there is a closed form of the sum?
    $endgroup$
    – SmileyCraft
    Dec 16 '18 at 12:54










  • $begingroup$
    Thanks, the first part of the question in a paper (which I doubt is wrong) asks to express $S_n$ in terms of $n$ before computing the limit of the quotient $dfrac{S_n}{n+1}$
    $endgroup$
    – Oussama Sarih
    Dec 16 '18 at 13:00








  • 2




    $begingroup$
    @OussamaSarih, can you link to the paper, or reproduce verbatim what it says about expressing $S_n$? Given that the first three $S_n$'s are $1/2$, $17/18$, and $985/738$, I'd be a little surprised to see any kind of simple closed formula. (Even well written papers make occasional mistakes.)
    $endgroup$
    – Barry Cipra
    Dec 16 '18 at 13:04












  • $begingroup$
    It's a homework sheet in french for highschool, I'll post a screenshot as soon as I get from one from the students who asked me to solve it.
    $endgroup$
    – Oussama Sarih
    Dec 16 '18 at 15:39














0












0








0


1



$begingroup$


I wonder whether there is a closed form for this sum
$$ S_n:=displaystylesum_{k=0}^n dfrac{4^k}{4^k+5^k}$$
The question asks to express the sum in terms of $n$ then to deduce the limit of $dfrac{S_n}{n+1}$.
I tried to use the following sum as an auxiliary sum
$$T_n:=displaystylesum_{k=0}^ndfrac{5^k}{4^k+5^k}$$
noticing that $S_n + T_n = n+1$.
Any thoughts about this ? thanks.






sequences-and-series summation telescopic-series







share|cite|improve this question









$endgroup$




I wonder whether there is a closed form for this sum
$$ S_n:=displaystylesum_{k=0}^n dfrac{4^k}{4^k+5^k}$$
The question asks to express the sum in terms of $n$ then to deduce the limit of $dfrac{S_n}{n+1}$.
I tried to use the following sum as an auxiliary sum
$$T_n:=displaystylesum_{k=0}^ndfrac{5^k}{4^k+5^k}$$
noticing that $S_n + T_n = n+1$.
Any thoughts about this ? thanks.






sequences-and-series summation telescopic-series




sequences-and-series summation telescopic-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 16 '18 at 12:38









Oussama SarihOussama Sarih

47827




47827








  • 4




    $begingroup$
    I don't know how to express the sum in terms of $n$, but I am a little confused. Don't we have $lim_{ntoinfty}S_n=sum_{k=0}^inftyfrac{4^k}{4^k+5^k}leqsum_{k=0}^infty(frac45)^k=frac1{1-frac45}=5$ and hence $lim_{ntoinfty}frac{S_n}{n+1}=0$?
    $endgroup$
    – SmileyCraft
    Dec 16 '18 at 12:49






  • 1




    $begingroup$
    Also, Wolfram Alpha gives a closed form involving the $q$-digamma function, which I've never heared of, and I doubt you have. What makes you think there is a closed form of the sum?
    $endgroup$
    – SmileyCraft
    Dec 16 '18 at 12:54










  • $begingroup$
    Thanks, the first part of the question in a paper (which I doubt is wrong) asks to express $S_n$ in terms of $n$ before computing the limit of the quotient $dfrac{S_n}{n+1}$
    $endgroup$
    – Oussama Sarih
    Dec 16 '18 at 13:00








  • 2




    $begingroup$
    @OussamaSarih, can you link to the paper, or reproduce verbatim what it says about expressing $S_n$? Given that the first three $S_n$'s are $1/2$, $17/18$, and $985/738$, I'd be a little surprised to see any kind of simple closed formula. (Even well written papers make occasional mistakes.)
    $endgroup$
    – Barry Cipra
    Dec 16 '18 at 13:04












  • $begingroup$
    It's a homework sheet in french for highschool, I'll post a screenshot as soon as I get from one from the students who asked me to solve it.
    $endgroup$
    – Oussama Sarih
    Dec 16 '18 at 15:39














  • 4




    $begingroup$
    I don't know how to express the sum in terms of $n$, but I am a little confused. Don't we have $lim_{ntoinfty}S_n=sum_{k=0}^inftyfrac{4^k}{4^k+5^k}leqsum_{k=0}^infty(frac45)^k=frac1{1-frac45}=5$ and hence $lim_{ntoinfty}frac{S_n}{n+1}=0$?
    $endgroup$
    – SmileyCraft
    Dec 16 '18 at 12:49






  • 1




    $begingroup$
    Also, Wolfram Alpha gives a closed form involving the $q$-digamma function, which I've never heared of, and I doubt you have. What makes you think there is a closed form of the sum?
    $endgroup$
    – SmileyCraft
    Dec 16 '18 at 12:54










  • $begingroup$
    Thanks, the first part of the question in a paper (which I doubt is wrong) asks to express $S_n$ in terms of $n$ before computing the limit of the quotient $dfrac{S_n}{n+1}$
    $endgroup$
    – Oussama Sarih
    Dec 16 '18 at 13:00








  • 2




    $begingroup$
    @OussamaSarih, can you link to the paper, or reproduce verbatim what it says about expressing $S_n$? Given that the first three $S_n$'s are $1/2$, $17/18$, and $985/738$, I'd be a little surprised to see any kind of simple closed formula. (Even well written papers make occasional mistakes.)
    $endgroup$
    – Barry Cipra
    Dec 16 '18 at 13:04












  • $begingroup$
    It's a homework sheet in french for highschool, I'll post a screenshot as soon as I get from one from the students who asked me to solve it.
    $endgroup$
    – Oussama Sarih
    Dec 16 '18 at 15:39








4




4




$begingroup$
I don't know how to express the sum in terms of $n$, but I am a little confused. Don't we have $lim_{ntoinfty}S_n=sum_{k=0}^inftyfrac{4^k}{4^k+5^k}leqsum_{k=0}^infty(frac45)^k=frac1{1-frac45}=5$ and hence $lim_{ntoinfty}frac{S_n}{n+1}=0$?
$endgroup$
– SmileyCraft
Dec 16 '18 at 12:49




$begingroup$
I don't know how to express the sum in terms of $n$, but I am a little confused. Don't we have $lim_{ntoinfty}S_n=sum_{k=0}^inftyfrac{4^k}{4^k+5^k}leqsum_{k=0}^infty(frac45)^k=frac1{1-frac45}=5$ and hence $lim_{ntoinfty}frac{S_n}{n+1}=0$?
$endgroup$
– SmileyCraft
Dec 16 '18 at 12:49




1




1




$begingroup$
Also, Wolfram Alpha gives a closed form involving the $q$-digamma function, which I've never heared of, and I doubt you have. What makes you think there is a closed form of the sum?
$endgroup$
– SmileyCraft
Dec 16 '18 at 12:54




$begingroup$
Also, Wolfram Alpha gives a closed form involving the $q$-digamma function, which I've never heared of, and I doubt you have. What makes you think there is a closed form of the sum?
$endgroup$
– SmileyCraft
Dec 16 '18 at 12:54












$begingroup$
Thanks, the first part of the question in a paper (which I doubt is wrong) asks to express $S_n$ in terms of $n$ before computing the limit of the quotient $dfrac{S_n}{n+1}$
$endgroup$
– Oussama Sarih
Dec 16 '18 at 13:00






$begingroup$
Thanks, the first part of the question in a paper (which I doubt is wrong) asks to express $S_n$ in terms of $n$ before computing the limit of the quotient $dfrac{S_n}{n+1}$
$endgroup$
– Oussama Sarih
Dec 16 '18 at 13:00






2




2




$begingroup$
@OussamaSarih, can you link to the paper, or reproduce verbatim what it says about expressing $S_n$? Given that the first three $S_n$'s are $1/2$, $17/18$, and $985/738$, I'd be a little surprised to see any kind of simple closed formula. (Even well written papers make occasional mistakes.)
$endgroup$
– Barry Cipra
Dec 16 '18 at 13:04






$begingroup$
@OussamaSarih, can you link to the paper, or reproduce verbatim what it says about expressing $S_n$? Given that the first three $S_n$'s are $1/2$, $17/18$, and $985/738$, I'd be a little surprised to see any kind of simple closed formula. (Even well written papers make occasional mistakes.)
$endgroup$
– Barry Cipra
Dec 16 '18 at 13:04














$begingroup$
It's a homework sheet in french for highschool, I'll post a screenshot as soon as I get from one from the students who asked me to solve it.
$endgroup$
– Oussama Sarih
Dec 16 '18 at 15:39




$begingroup$
It's a homework sheet in french for highschool, I'll post a screenshot as soon as I get from one from the students who asked me to solve it.
$endgroup$
– Oussama Sarih
Dec 16 '18 at 15:39










1 Answer
1






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oldest

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1












$begingroup$

As hinted by SmileyCraft, I cannot think of any simple closed form to express $S_n$.



My best guess is to use Big $mathcal{O}$ notation. As you thought, $S_n = n+1 - T_n$ then write $T_n = sum_{k=0}^n u_k$ with $u_k = frac{1}{1+left(frac{4}{5}right)^k}underset{ktoinfty}{=}1 + mathcal{O}left(frac{4}{5}right)^k$. Now $left(frac{4}{5}right)^k$ is a positive real sequence thus sommable in Big $mathcal{O}$ notation.



This yields:
$$
S_n underset{nto infty}{=} n+1 - left[(n+1) + mathcal{O}(1) right] = mathcal{O}(1)
$$



since $sum_{k=0}^n left(frac{4}{5}right)^k underset{nto infty}{=} mathcal{O}(1)$.



Then $frac{S_n}{n+1} = mathcal{O}(frac{1}{n+1})$.



Note that it is very similar and perfectly equivalent to what SmileyCraft does but it does give you an expression of $S_n$ (though trivial) depending on $n$.






share|cite|improve this answer









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    $begingroup$

    As hinted by SmileyCraft, I cannot think of any simple closed form to express $S_n$.



    My best guess is to use Big $mathcal{O}$ notation. As you thought, $S_n = n+1 - T_n$ then write $T_n = sum_{k=0}^n u_k$ with $u_k = frac{1}{1+left(frac{4}{5}right)^k}underset{ktoinfty}{=}1 + mathcal{O}left(frac{4}{5}right)^k$. Now $left(frac{4}{5}right)^k$ is a positive real sequence thus sommable in Big $mathcal{O}$ notation.



    This yields:
    $$
    S_n underset{nto infty}{=} n+1 - left[(n+1) + mathcal{O}(1) right] = mathcal{O}(1)
    $$



    since $sum_{k=0}^n left(frac{4}{5}right)^k underset{nto infty}{=} mathcal{O}(1)$.



    Then $frac{S_n}{n+1} = mathcal{O}(frac{1}{n+1})$.



    Note that it is very similar and perfectly equivalent to what SmileyCraft does but it does give you an expression of $S_n$ (though trivial) depending on $n$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      As hinted by SmileyCraft, I cannot think of any simple closed form to express $S_n$.



      My best guess is to use Big $mathcal{O}$ notation. As you thought, $S_n = n+1 - T_n$ then write $T_n = sum_{k=0}^n u_k$ with $u_k = frac{1}{1+left(frac{4}{5}right)^k}underset{ktoinfty}{=}1 + mathcal{O}left(frac{4}{5}right)^k$. Now $left(frac{4}{5}right)^k$ is a positive real sequence thus sommable in Big $mathcal{O}$ notation.



      This yields:
      $$
      S_n underset{nto infty}{=} n+1 - left[(n+1) + mathcal{O}(1) right] = mathcal{O}(1)
      $$



      since $sum_{k=0}^n left(frac{4}{5}right)^k underset{nto infty}{=} mathcal{O}(1)$.



      Then $frac{S_n}{n+1} = mathcal{O}(frac{1}{n+1})$.



      Note that it is very similar and perfectly equivalent to what SmileyCraft does but it does give you an expression of $S_n$ (though trivial) depending on $n$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        As hinted by SmileyCraft, I cannot think of any simple closed form to express $S_n$.



        My best guess is to use Big $mathcal{O}$ notation. As you thought, $S_n = n+1 - T_n$ then write $T_n = sum_{k=0}^n u_k$ with $u_k = frac{1}{1+left(frac{4}{5}right)^k}underset{ktoinfty}{=}1 + mathcal{O}left(frac{4}{5}right)^k$. Now $left(frac{4}{5}right)^k$ is a positive real sequence thus sommable in Big $mathcal{O}$ notation.



        This yields:
        $$
        S_n underset{nto infty}{=} n+1 - left[(n+1) + mathcal{O}(1) right] = mathcal{O}(1)
        $$



        since $sum_{k=0}^n left(frac{4}{5}right)^k underset{nto infty}{=} mathcal{O}(1)$.



        Then $frac{S_n}{n+1} = mathcal{O}(frac{1}{n+1})$.



        Note that it is very similar and perfectly equivalent to what SmileyCraft does but it does give you an expression of $S_n$ (though trivial) depending on $n$.






        share|cite|improve this answer









        $endgroup$



        As hinted by SmileyCraft, I cannot think of any simple closed form to express $S_n$.



        My best guess is to use Big $mathcal{O}$ notation. As you thought, $S_n = n+1 - T_n$ then write $T_n = sum_{k=0}^n u_k$ with $u_k = frac{1}{1+left(frac{4}{5}right)^k}underset{ktoinfty}{=}1 + mathcal{O}left(frac{4}{5}right)^k$. Now $left(frac{4}{5}right)^k$ is a positive real sequence thus sommable in Big $mathcal{O}$ notation.



        This yields:
        $$
        S_n underset{nto infty}{=} n+1 - left[(n+1) + mathcal{O}(1) right] = mathcal{O}(1)
        $$



        since $sum_{k=0}^n left(frac{4}{5}right)^k underset{nto infty}{=} mathcal{O}(1)$.



        Then $frac{S_n}{n+1} = mathcal{O}(frac{1}{n+1})$.



        Note that it is very similar and perfectly equivalent to what SmileyCraft does but it does give you an expression of $S_n$ (though trivial) depending on $n$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 16 '18 at 13:27









        Bill O'HaranBill O'Haran

        2,5431418




        2,5431418






























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