Finding a upper bound (estimation) for the infinite series:...












0












$begingroup$


I have the infinite sum:
begin{align}
S_n = frac{n^5!}{(n+1)^5!}+frac{n^5!}{(n+2)^5!}+frac{n^5!}{(n+3)^5!}+cdots
end{align}
I must find a upper bound for $S_n$, that is, $0<S_n<f(n)$, where $f(n)$ is a function that $lim_{ntoinfty} f(n) = 0$, because, I'm not sure if this is allowed in a infite series, but $n^5!/(n+1)^5!$ goes to zero as $n$ tends to infinite, and the same happens for all other terms, so $lim_{nto infty}S_n = 0 ::(?)$. So exists a $f(n)$ such that, $0<S_n<f(n)$ where $f(n)to 0 $ as $nto infty$




For example: try to estimate (find a upper bound) the infinite series: $R_n = n!/(n+1)! + n!/(n+2)!+cdots$




I can cancel the terms after expanding the factorial, resulting in $R_n = 1/(n+1)+1/(n+2)(n+1)+cdots<1/(n+1)+1/(n+1)(n+1)+cdots = 1/n$ (because it's a geometric series). So we have that $0<R_n<1/n$. And that's enough. Now I must do the same for $S_n$ defined above, but I can't cancel terms after expanding the factorial.






sequences-and-series estimation upper-lower-bounds







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  • $begingroup$
    "Upper bound" or "estimate"? Not the same thing...
    $endgroup$
    – Did
    Jan 6 at 20:53
















0












$begingroup$


I have the infinite sum:
begin{align}
S_n = frac{n^5!}{(n+1)^5!}+frac{n^5!}{(n+2)^5!}+frac{n^5!}{(n+3)^5!}+cdots
end{align}
I must find a upper bound for $S_n$, that is, $0<S_n<f(n)$, where $f(n)$ is a function that $lim_{ntoinfty} f(n) = 0$, because, I'm not sure if this is allowed in a infite series, but $n^5!/(n+1)^5!$ goes to zero as $n$ tends to infinite, and the same happens for all other terms, so $lim_{nto infty}S_n = 0 ::(?)$. So exists a $f(n)$ such that, $0<S_n<f(n)$ where $f(n)to 0 $ as $nto infty$




For example: try to estimate (find a upper bound) the infinite series: $R_n = n!/(n+1)! + n!/(n+2)!+cdots$




I can cancel the terms after expanding the factorial, resulting in $R_n = 1/(n+1)+1/(n+2)(n+1)+cdots<1/(n+1)+1/(n+1)(n+1)+cdots = 1/n$ (because it's a geometric series). So we have that $0<R_n<1/n$. And that's enough. Now I must do the same for $S_n$ defined above, but I can't cancel terms after expanding the factorial.






sequences-and-series estimation upper-lower-bounds







share|cite|improve this question











$endgroup$












  • $begingroup$
    "Upper bound" or "estimate"? Not the same thing...
    $endgroup$
    – Did
    Jan 6 at 20:53














0












0








0





$begingroup$


I have the infinite sum:
begin{align}
S_n = frac{n^5!}{(n+1)^5!}+frac{n^5!}{(n+2)^5!}+frac{n^5!}{(n+3)^5!}+cdots
end{align}
I must find a upper bound for $S_n$, that is, $0<S_n<f(n)$, where $f(n)$ is a function that $lim_{ntoinfty} f(n) = 0$, because, I'm not sure if this is allowed in a infite series, but $n^5!/(n+1)^5!$ goes to zero as $n$ tends to infinite, and the same happens for all other terms, so $lim_{nto infty}S_n = 0 ::(?)$. So exists a $f(n)$ such that, $0<S_n<f(n)$ where $f(n)to 0 $ as $nto infty$




For example: try to estimate (find a upper bound) the infinite series: $R_n = n!/(n+1)! + n!/(n+2)!+cdots$




I can cancel the terms after expanding the factorial, resulting in $R_n = 1/(n+1)+1/(n+2)(n+1)+cdots<1/(n+1)+1/(n+1)(n+1)+cdots = 1/n$ (because it's a geometric series). So we have that $0<R_n<1/n$. And that's enough. Now I must do the same for $S_n$ defined above, but I can't cancel terms after expanding the factorial.






sequences-and-series estimation upper-lower-bounds







share|cite|improve this question











$endgroup$




I have the infinite sum:
begin{align}
S_n = frac{n^5!}{(n+1)^5!}+frac{n^5!}{(n+2)^5!}+frac{n^5!}{(n+3)^5!}+cdots
end{align}
I must find a upper bound for $S_n$, that is, $0<S_n<f(n)$, where $f(n)$ is a function that $lim_{ntoinfty} f(n) = 0$, because, I'm not sure if this is allowed in a infite series, but $n^5!/(n+1)^5!$ goes to zero as $n$ tends to infinite, and the same happens for all other terms, so $lim_{nto infty}S_n = 0 ::(?)$. So exists a $f(n)$ such that, $0<S_n<f(n)$ where $f(n)to 0 $ as $nto infty$




For example: try to estimate (find a upper bound) the infinite series: $R_n = n!/(n+1)! + n!/(n+2)!+cdots$




I can cancel the terms after expanding the factorial, resulting in $R_n = 1/(n+1)+1/(n+2)(n+1)+cdots<1/(n+1)+1/(n+1)(n+1)+cdots = 1/n$ (because it's a geometric series). So we have that $0<R_n<1/n$. And that's enough. Now I must do the same for $S_n$ defined above, but I can't cancel terms after expanding the factorial.






sequences-and-series estimation upper-lower-bounds




sequences-and-series estimation upper-lower-bounds






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edited Jan 6 at 20:41









Bernard

123k741117




123k741117










asked Jan 6 at 19:39









PintecoPinteco

800313




800313












  • $begingroup$
    "Upper bound" or "estimate"? Not the same thing...
    $endgroup$
    – Did
    Jan 6 at 20:53


















  • $begingroup$
    "Upper bound" or "estimate"? Not the same thing...
    $endgroup$
    – Did
    Jan 6 at 20:53
















$begingroup$
"Upper bound" or "estimate"? Not the same thing...
$endgroup$
– Did
Jan 6 at 20:53




$begingroup$
"Upper bound" or "estimate"? Not the same thing...
$endgroup$
– Did
Jan 6 at 20:53










1 Answer
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$begingroup$

It holds that
$$
frac{n^5!}{(n+k)^5!}lefrac{1}{(n^5+k)(n^5+k+1)}=frac{1}{n^5+k}-frac{1}{n^5+k+1}
$$ since $(n+k)^5ge n+k+1$, and hence
$$
S_n = sum_{k=1}^infty frac{n^5!}{(n+k)^5!}le sum_{k=1}^infty left(frac{1}{n^5+k}-frac{1}{n^5+k+1}right)=frac{1}{n^5+1}to 0.
$$






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    $begingroup$

    It holds that
    $$
    frac{n^5!}{(n+k)^5!}lefrac{1}{(n^5+k)(n^5+k+1)}=frac{1}{n^5+k}-frac{1}{n^5+k+1}
    $$     since $(n+k)^5ge n+k+1$, and hence
    $$
    S_n = sum_{k=1}^infty frac{n^5!}{(n+k)^5!}le sum_{k=1}^infty left(frac{1}{n^5+k}-frac{1}{n^5+k+1}right)=frac{1}{n^5+1}to 0.
    $$






    share|cite|improve this answer











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      3












      $begingroup$

      It holds that
      $$
      frac{n^5!}{(n+k)^5!}lefrac{1}{(n^5+k)(n^5+k+1)}=frac{1}{n^5+k}-frac{1}{n^5+k+1}
      $$       since $(n+k)^5ge n+k+1$, and hence
      $$
      S_n = sum_{k=1}^infty frac{n^5!}{(n+k)^5!}le sum_{k=1}^infty left(frac{1}{n^5+k}-frac{1}{n^5+k+1}right)=frac{1}{n^5+1}to 0.
      $$






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        It holds that
        $$
        frac{n^5!}{(n+k)^5!}lefrac{1}{(n^5+k)(n^5+k+1)}=frac{1}{n^5+k}-frac{1}{n^5+k+1}
        $$         since $(n+k)^5ge n+k+1$, and hence
        $$
        S_n = sum_{k=1}^infty frac{n^5!}{(n+k)^5!}le sum_{k=1}^infty left(frac{1}{n^5+k}-frac{1}{n^5+k+1}right)=frac{1}{n^5+1}to 0.
        $$






        share|cite|improve this answer











        $endgroup$



        It holds that
        $$
        frac{n^5!}{(n+k)^5!}lefrac{1}{(n^5+k)(n^5+k+1)}=frac{1}{n^5+k}-frac{1}{n^5+k+1}
        $$         since $(n+k)^5ge n+k+1$, and hence
        $$
        S_n = sum_{k=1}^infty frac{n^5!}{(n+k)^5!}le sum_{k=1}^infty left(frac{1}{n^5+k}-frac{1}{n^5+k+1}right)=frac{1}{n^5+1}to 0.
        $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 6 at 21:37

























        answered Jan 6 at 19:53









        SongSong

        18.5k21651




        18.5k21651






























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