uniformly continuous function $f$ such that $sum 1/f(n)$ is convergent?












4












$begingroup$


Does there exist a uniformly continuous function $f:[1,infty)to mathbb R$ such that $sum_{n=1}^infty 1/f(n)$ is convergent ?



I know that $exists M>0$ such that $|f(x)|< Mx, forall xin [1,infty)$, so $|1/f(n)|>1/(Mn) ,forall n ge 1$, thus $sum_{n=1}^infty |1/f(n)|$ is divergent. But I don't know what happens with $sum_{n=1}^infty 1/f(n)$.



Please help






real-analysis sequences-and-series uniform-continuity







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hint: can you figure out how to find an elementary $f$ that alternates in sign at each integer, is increasing in absolute value, and grows at $o(x)$ as $xtoinfty$? (Bigger hint: can you figure out how to take a positive function that satisfies the last two conditions, and make it satisfy the first?)
    $endgroup$
    – Steven Stadnicki
    Dec 11 '18 at 21:29












  • $begingroup$
    @StevenStadnicki: not really no ... I have no idea what you're talking about ...
    $endgroup$
    – user521337
    Dec 11 '18 at 21:32










  • $begingroup$
    Try this: can you think of an elementary function whose value at $n$ is $(-1)^n$?
    $endgroup$
    – Steven Stadnicki
    Dec 11 '18 at 21:35






  • 4




    $begingroup$
    Product of two uniformly continuous functions is not necessarily uniformly continuous...they must be bounded for that to be true. I wouldn't trust their suggestions.
    $endgroup$
    – Jeffery Opoku-Mensah
    Dec 11 '18 at 21:56






  • 1




    $begingroup$
    @JefferyOpoku-Mensah You're right. I followed the line of comments without thinking. It's not immediate it works. (And won't for every such choice of functions)
    $endgroup$
    – Clement C.
    Dec 11 '18 at 22:02


















4












$begingroup$


Does there exist a uniformly continuous function $f:[1,infty)to mathbb R$ such that $sum_{n=1}^infty 1/f(n)$ is convergent ?



I know that $exists M>0$ such that $|f(x)|< Mx, forall xin [1,infty)$, so $|1/f(n)|>1/(Mn) ,forall n ge 1$, thus $sum_{n=1}^infty |1/f(n)|$ is divergent. But I don't know what happens with $sum_{n=1}^infty 1/f(n)$.



Please help






real-analysis sequences-and-series uniform-continuity







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hint: can you figure out how to find an elementary $f$ that alternates in sign at each integer, is increasing in absolute value, and grows at $o(x)$ as $xtoinfty$? (Bigger hint: can you figure out how to take a positive function that satisfies the last two conditions, and make it satisfy the first?)
    $endgroup$
    – Steven Stadnicki
    Dec 11 '18 at 21:29












  • $begingroup$
    @StevenStadnicki: not really no ... I have no idea what you're talking about ...
    $endgroup$
    – user521337
    Dec 11 '18 at 21:32










  • $begingroup$
    Try this: can you think of an elementary function whose value at $n$ is $(-1)^n$?
    $endgroup$
    – Steven Stadnicki
    Dec 11 '18 at 21:35






  • 4




    $begingroup$
    Product of two uniformly continuous functions is not necessarily uniformly continuous...they must be bounded for that to be true. I wouldn't trust their suggestions.
    $endgroup$
    – Jeffery Opoku-Mensah
    Dec 11 '18 at 21:56






  • 1




    $begingroup$
    @JefferyOpoku-Mensah You're right. I followed the line of comments without thinking. It's not immediate it works. (And won't for every such choice of functions)
    $endgroup$
    – Clement C.
    Dec 11 '18 at 22:02
















4












4








4





$begingroup$


Does there exist a uniformly continuous function $f:[1,infty)to mathbb R$ such that $sum_{n=1}^infty 1/f(n)$ is convergent ?



I know that $exists M>0$ such that $|f(x)|< Mx, forall xin [1,infty)$, so $|1/f(n)|>1/(Mn) ,forall n ge 1$, thus $sum_{n=1}^infty |1/f(n)|$ is divergent. But I don't know what happens with $sum_{n=1}^infty 1/f(n)$.



Please help






real-analysis sequences-and-series uniform-continuity







share|cite|improve this question









$endgroup$




Does there exist a uniformly continuous function $f:[1,infty)to mathbb R$ such that $sum_{n=1}^infty 1/f(n)$ is convergent ?



I know that $exists M>0$ such that $|f(x)|< Mx, forall xin [1,infty)$, so $|1/f(n)|>1/(Mn) ,forall n ge 1$, thus $sum_{n=1}^infty |1/f(n)|$ is divergent. But I don't know what happens with $sum_{n=1}^infty 1/f(n)$.



Please help






real-analysis sequences-and-series uniform-continuity




real-analysis sequences-and-series uniform-continuity






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 11 '18 at 21:23









user521337user521337

1,2041417




1,2041417








  • 1




    $begingroup$
    Hint: can you figure out how to find an elementary $f$ that alternates in sign at each integer, is increasing in absolute value, and grows at $o(x)$ as $xtoinfty$? (Bigger hint: can you figure out how to take a positive function that satisfies the last two conditions, and make it satisfy the first?)
    $endgroup$
    – Steven Stadnicki
    Dec 11 '18 at 21:29












  • $begingroup$
    @StevenStadnicki: not really no ... I have no idea what you're talking about ...
    $endgroup$
    – user521337
    Dec 11 '18 at 21:32










  • $begingroup$
    Try this: can you think of an elementary function whose value at $n$ is $(-1)^n$?
    $endgroup$
    – Steven Stadnicki
    Dec 11 '18 at 21:35






  • 4




    $begingroup$
    Product of two uniformly continuous functions is not necessarily uniformly continuous...they must be bounded for that to be true. I wouldn't trust their suggestions.
    $endgroup$
    – Jeffery Opoku-Mensah
    Dec 11 '18 at 21:56






  • 1




    $begingroup$
    @JefferyOpoku-Mensah You're right. I followed the line of comments without thinking. It's not immediate it works. (And won't for every such choice of functions)
    $endgroup$
    – Clement C.
    Dec 11 '18 at 22:02
















  • 1




    $begingroup$
    Hint: can you figure out how to find an elementary $f$ that alternates in sign at each integer, is increasing in absolute value, and grows at $o(x)$ as $xtoinfty$? (Bigger hint: can you figure out how to take a positive function that satisfies the last two conditions, and make it satisfy the first?)
    $endgroup$
    – Steven Stadnicki
    Dec 11 '18 at 21:29












  • $begingroup$
    @StevenStadnicki: not really no ... I have no idea what you're talking about ...
    $endgroup$
    – user521337
    Dec 11 '18 at 21:32










  • $begingroup$
    Try this: can you think of an elementary function whose value at $n$ is $(-1)^n$?
    $endgroup$
    – Steven Stadnicki
    Dec 11 '18 at 21:35






  • 4




    $begingroup$
    Product of two uniformly continuous functions is not necessarily uniformly continuous...they must be bounded for that to be true. I wouldn't trust their suggestions.
    $endgroup$
    – Jeffery Opoku-Mensah
    Dec 11 '18 at 21:56






  • 1




    $begingroup$
    @JefferyOpoku-Mensah You're right. I followed the line of comments without thinking. It's not immediate it works. (And won't for every such choice of functions)
    $endgroup$
    – Clement C.
    Dec 11 '18 at 22:02










1




1




$begingroup$
Hint: can you figure out how to find an elementary $f$ that alternates in sign at each integer, is increasing in absolute value, and grows at $o(x)$ as $xtoinfty$? (Bigger hint: can you figure out how to take a positive function that satisfies the last two conditions, and make it satisfy the first?)
$endgroup$
– Steven Stadnicki
Dec 11 '18 at 21:29






$begingroup$
Hint: can you figure out how to find an elementary $f$ that alternates in sign at each integer, is increasing in absolute value, and grows at $o(x)$ as $xtoinfty$? (Bigger hint: can you figure out how to take a positive function that satisfies the last two conditions, and make it satisfy the first?)
$endgroup$
– Steven Stadnicki
Dec 11 '18 at 21:29














$begingroup$
@StevenStadnicki: not really no ... I have no idea what you're talking about ...
$endgroup$
– user521337
Dec 11 '18 at 21:32




$begingroup$
@StevenStadnicki: not really no ... I have no idea what you're talking about ...
$endgroup$
– user521337
Dec 11 '18 at 21:32












$begingroup$
Try this: can you think of an elementary function whose value at $n$ is $(-1)^n$?
$endgroup$
– Steven Stadnicki
Dec 11 '18 at 21:35




$begingroup$
Try this: can you think of an elementary function whose value at $n$ is $(-1)^n$?
$endgroup$
– Steven Stadnicki
Dec 11 '18 at 21:35




4




4




$begingroup$
Product of two uniformly continuous functions is not necessarily uniformly continuous...they must be bounded for that to be true. I wouldn't trust their suggestions.
$endgroup$
– Jeffery Opoku-Mensah
Dec 11 '18 at 21:56




$begingroup$
Product of two uniformly continuous functions is not necessarily uniformly continuous...they must be bounded for that to be true. I wouldn't trust their suggestions.
$endgroup$
– Jeffery Opoku-Mensah
Dec 11 '18 at 21:56




1




1




$begingroup$
@JefferyOpoku-Mensah You're right. I followed the line of comments without thinking. It's not immediate it works. (And won't for every such choice of functions)
$endgroup$
– Clement C.
Dec 11 '18 at 22:02






$begingroup$
@JefferyOpoku-Mensah You're right. I followed the line of comments without thinking. It's not immediate it works. (And won't for every such choice of functions)
$endgroup$
– Clement C.
Dec 11 '18 at 22:02












2 Answers
2






active

oldest

votes


















5












$begingroup$

Let us assume that such a function $f$ exists. From the uniform continuity we get a constant $M > 0$ such that
$$ tag 1
|f(n+1) - f(n)| < M
$$
for all $n in Bbb N$. In particular $|f(n)| < |f(1)| + nM$, so that
$sum_{n=1}^infty frac{1}{|f(n)|}$ diverges (as you already observed).



If $sum_{n=1}^infty frac{1}{f(n)}$ is convergent then
necessarily $|f(n)| to infty$ so that
$$ tag 2
|f(n)| > M
$$
for $n ge n_0$.



Combining these inequalities it follows that for $n ge n_0$, all $f(n)$ have the same sign, so that $sum_{n=1}^infty frac{1}{f(n)}$
is absolutely convergent, in contradiction to the above observation.



Therefore no such function $f$ exists.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    You should say $|f(n)| to infty $ ...
    $endgroup$
    – user521337
    Dec 11 '18 at 22:20












  • $begingroup$
    I don't see why $f(n)$ s ultimately have the same sign ... could you please elaborate on that ?
    $endgroup$
    – user521337
    Dec 11 '18 at 22:23






  • 1




    $begingroup$
    @user521337: If $f(n)$ and $f(n+1)$ have different sign and both have an absolute value $> M$ then their absolute difference would be at least $2M$.
    $endgroup$
    – Martin R
    Dec 11 '18 at 22:25





















1












$begingroup$

Assume that $sum 1/f$ converges. Then for some $N_k in mathbb{N}$,
$$n > N_k implies |1/f(n)| < 1/k implies |f(n)| > k.$$



$f$ is uniformly continuous. Therefore, there exists $m in mathbb{N}$ such that $|x-y| < 1/m implies |f(x)-f(y)| < 1$. By the triangle inequality, this implies $|f(n+1)-f(n)|<m$ for all $n$.



Then if $a, a+1 > N_{lceil m/2 rceil} = N'$ are integers such that WLOG $f(a) > 0$ and $f(a+1) < 0$, then $f(a) > k$ and $f(a+1) < -k$. Then $|f(a+1)-f(a)|> 2lceil m/2 rceil > m$, contradiction. Therefore, $f(n)$ must not change sign for integer $n > N'$.



So, assume WLOG that $f(n)$ is eventually positive. Note this implies for $n > N'$ and $r in mathbb{N}$, that $f(n+r) > f(n)+ rm$.



Write
$$sum_{n=1}^{infty} frac{1}{f(n)} = sum_{n=1}^{N'} frac{1}{f(n)} + sum_{n= N'+1}^{infty} frac{1}{f(n)}$$



Realize that for the following series whose terms are strictly positive,
$$sum_{n= N'+1}^{infty} frac{1}{f(n)} > sum_{n = 0}^{infty} frac{1}{f(N'+1) + nm},$$



and the RHS obviously diverges. Contradiction.






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









    5












    $begingroup$

    Let us assume that such a function $f$ exists. From the uniform continuity we get a constant $M > 0$ such that
    $$ tag 1
    |f(n+1) - f(n)| < M
    $$
    for all $n in Bbb N$. In particular $|f(n)| < |f(1)| + nM$, so that
    $sum_{n=1}^infty frac{1}{|f(n)|}$ diverges (as you already observed).



    If $sum_{n=1}^infty frac{1}{f(n)}$ is convergent then
    necessarily $|f(n)| to infty$ so that
    $$ tag 2
    |f(n)| > M
    $$
    for $n ge n_0$.



    Combining these inequalities it follows that for $n ge n_0$, all $f(n)$ have the same sign, so that $sum_{n=1}^infty frac{1}{f(n)}$
    is absolutely convergent, in contradiction to the above observation.



    Therefore no such function $f$ exists.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      You should say $|f(n)| to infty $ ...
      $endgroup$
      – user521337
      Dec 11 '18 at 22:20












    • $begingroup$
      I don't see why $f(n)$ s ultimately have the same sign ... could you please elaborate on that ?
      $endgroup$
      – user521337
      Dec 11 '18 at 22:23






    • 1




      $begingroup$
      @user521337: If $f(n)$ and $f(n+1)$ have different sign and both have an absolute value $> M$ then their absolute difference would be at least $2M$.
      $endgroup$
      – Martin R
      Dec 11 '18 at 22:25


















    5












    $begingroup$

    Let us assume that such a function $f$ exists. From the uniform continuity we get a constant $M > 0$ such that
    $$ tag 1
    |f(n+1) - f(n)| < M
    $$
    for all $n in Bbb N$. In particular $|f(n)| < |f(1)| + nM$, so that
    $sum_{n=1}^infty frac{1}{|f(n)|}$ diverges (as you already observed).



    If $sum_{n=1}^infty frac{1}{f(n)}$ is convergent then
    necessarily $|f(n)| to infty$ so that
    $$ tag 2
    |f(n)| > M
    $$
    for $n ge n_0$.



    Combining these inequalities it follows that for $n ge n_0$, all $f(n)$ have the same sign, so that $sum_{n=1}^infty frac{1}{f(n)}$
    is absolutely convergent, in contradiction to the above observation.



    Therefore no such function $f$ exists.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      You should say $|f(n)| to infty $ ...
      $endgroup$
      – user521337
      Dec 11 '18 at 22:20












    • $begingroup$
      I don't see why $f(n)$ s ultimately have the same sign ... could you please elaborate on that ?
      $endgroup$
      – user521337
      Dec 11 '18 at 22:23






    • 1




      $begingroup$
      @user521337: If $f(n)$ and $f(n+1)$ have different sign and both have an absolute value $> M$ then their absolute difference would be at least $2M$.
      $endgroup$
      – Martin R
      Dec 11 '18 at 22:25
















    5












    5








    5





    $begingroup$

    Let us assume that such a function $f$ exists. From the uniform continuity we get a constant $M > 0$ such that
    $$ tag 1
    |f(n+1) - f(n)| < M
    $$
    for all $n in Bbb N$. In particular $|f(n)| < |f(1)| + nM$, so that
    $sum_{n=1}^infty frac{1}{|f(n)|}$ diverges (as you already observed).



    If $sum_{n=1}^infty frac{1}{f(n)}$ is convergent then
    necessarily $|f(n)| to infty$ so that
    $$ tag 2
    |f(n)| > M
    $$
    for $n ge n_0$.



    Combining these inequalities it follows that for $n ge n_0$, all $f(n)$ have the same sign, so that $sum_{n=1}^infty frac{1}{f(n)}$
    is absolutely convergent, in contradiction to the above observation.



    Therefore no such function $f$ exists.






    share|cite|improve this answer











    $endgroup$



    Let us assume that such a function $f$ exists. From the uniform continuity we get a constant $M > 0$ such that
    $$ tag 1
    |f(n+1) - f(n)| < M
    $$
    for all $n in Bbb N$. In particular $|f(n)| < |f(1)| + nM$, so that
    $sum_{n=1}^infty frac{1}{|f(n)|}$ diverges (as you already observed).



    If $sum_{n=1}^infty frac{1}{f(n)}$ is convergent then
    necessarily $|f(n)| to infty$ so that
    $$ tag 2
    |f(n)| > M
    $$
    for $n ge n_0$.



    Combining these inequalities it follows that for $n ge n_0$, all $f(n)$ have the same sign, so that $sum_{n=1}^infty frac{1}{f(n)}$
    is absolutely convergent, in contradiction to the above observation.



    Therefore no such function $f$ exists.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 11 '18 at 22:28

























    answered Dec 11 '18 at 22:15









    Martin RMartin R

    30.2k33558




    30.2k33558








    • 1




      $begingroup$
      You should say $|f(n)| to infty $ ...
      $endgroup$
      – user521337
      Dec 11 '18 at 22:20












    • $begingroup$
      I don't see why $f(n)$ s ultimately have the same sign ... could you please elaborate on that ?
      $endgroup$
      – user521337
      Dec 11 '18 at 22:23






    • 1




      $begingroup$
      @user521337: If $f(n)$ and $f(n+1)$ have different sign and both have an absolute value $> M$ then their absolute difference would be at least $2M$.
      $endgroup$
      – Martin R
      Dec 11 '18 at 22:25
















    • 1




      $begingroup$
      You should say $|f(n)| to infty $ ...
      $endgroup$
      – user521337
      Dec 11 '18 at 22:20












    • $begingroup$
      I don't see why $f(n)$ s ultimately have the same sign ... could you please elaborate on that ?
      $endgroup$
      – user521337
      Dec 11 '18 at 22:23






    • 1




      $begingroup$
      @user521337: If $f(n)$ and $f(n+1)$ have different sign and both have an absolute value $> M$ then their absolute difference would be at least $2M$.
      $endgroup$
      – Martin R
      Dec 11 '18 at 22:25










    1




    1




    $begingroup$
    You should say $|f(n)| to infty $ ...
    $endgroup$
    – user521337
    Dec 11 '18 at 22:20






    $begingroup$
    You should say $|f(n)| to infty $ ...
    $endgroup$
    – user521337
    Dec 11 '18 at 22:20














    $begingroup$
    I don't see why $f(n)$ s ultimately have the same sign ... could you please elaborate on that ?
    $endgroup$
    – user521337
    Dec 11 '18 at 22:23




    $begingroup$
    I don't see why $f(n)$ s ultimately have the same sign ... could you please elaborate on that ?
    $endgroup$
    – user521337
    Dec 11 '18 at 22:23




    1




    1




    $begingroup$
    @user521337: If $f(n)$ and $f(n+1)$ have different sign and both have an absolute value $> M$ then their absolute difference would be at least $2M$.
    $endgroup$
    – Martin R
    Dec 11 '18 at 22:25






    $begingroup$
    @user521337: If $f(n)$ and $f(n+1)$ have different sign and both have an absolute value $> M$ then their absolute difference would be at least $2M$.
    $endgroup$
    – Martin R
    Dec 11 '18 at 22:25













    1












    $begingroup$

    Assume that $sum 1/f$ converges. Then for some $N_k in mathbb{N}$,
    $$n > N_k implies |1/f(n)| < 1/k implies |f(n)| > k.$$



    $f$ is uniformly continuous. Therefore, there exists $m in mathbb{N}$ such that $|x-y| < 1/m implies |f(x)-f(y)| < 1$. By the triangle inequality, this implies $|f(n+1)-f(n)|<m$ for all $n$.



    Then if $a, a+1 > N_{lceil m/2 rceil} = N'$ are integers such that WLOG $f(a) > 0$ and $f(a+1) < 0$, then $f(a) > k$ and $f(a+1) < -k$. Then $|f(a+1)-f(a)|> 2lceil m/2 rceil > m$, contradiction. Therefore, $f(n)$ must not change sign for integer $n > N'$.



    So, assume WLOG that $f(n)$ is eventually positive. Note this implies for $n > N'$ and $r in mathbb{N}$, that $f(n+r) > f(n)+ rm$.



    Write
    $$sum_{n=1}^{infty} frac{1}{f(n)} = sum_{n=1}^{N'} frac{1}{f(n)} + sum_{n= N'+1}^{infty} frac{1}{f(n)}$$



    Realize that for the following series whose terms are strictly positive,
    $$sum_{n= N'+1}^{infty} frac{1}{f(n)} > sum_{n = 0}^{infty} frac{1}{f(N'+1) + nm},$$



    and the RHS obviously diverges. Contradiction.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Assume that $sum 1/f$ converges. Then for some $N_k in mathbb{N}$,
      $$n > N_k implies |1/f(n)| < 1/k implies |f(n)| > k.$$



      $f$ is uniformly continuous. Therefore, there exists $m in mathbb{N}$ such that $|x-y| < 1/m implies |f(x)-f(y)| < 1$. By the triangle inequality, this implies $|f(n+1)-f(n)|<m$ for all $n$.



      Then if $a, a+1 > N_{lceil m/2 rceil} = N'$ are integers such that WLOG $f(a) > 0$ and $f(a+1) < 0$, then $f(a) > k$ and $f(a+1) < -k$. Then $|f(a+1)-f(a)|> 2lceil m/2 rceil > m$, contradiction. Therefore, $f(n)$ must not change sign for integer $n > N'$.



      So, assume WLOG that $f(n)$ is eventually positive. Note this implies for $n > N'$ and $r in mathbb{N}$, that $f(n+r) > f(n)+ rm$.



      Write
      $$sum_{n=1}^{infty} frac{1}{f(n)} = sum_{n=1}^{N'} frac{1}{f(n)} + sum_{n= N'+1}^{infty} frac{1}{f(n)}$$



      Realize that for the following series whose terms are strictly positive,
      $$sum_{n= N'+1}^{infty} frac{1}{f(n)} > sum_{n = 0}^{infty} frac{1}{f(N'+1) + nm},$$



      and the RHS obviously diverges. Contradiction.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Assume that $sum 1/f$ converges. Then for some $N_k in mathbb{N}$,
        $$n > N_k implies |1/f(n)| < 1/k implies |f(n)| > k.$$



        $f$ is uniformly continuous. Therefore, there exists $m in mathbb{N}$ such that $|x-y| < 1/m implies |f(x)-f(y)| < 1$. By the triangle inequality, this implies $|f(n+1)-f(n)|<m$ for all $n$.



        Then if $a, a+1 > N_{lceil m/2 rceil} = N'$ are integers such that WLOG $f(a) > 0$ and $f(a+1) < 0$, then $f(a) > k$ and $f(a+1) < -k$. Then $|f(a+1)-f(a)|> 2lceil m/2 rceil > m$, contradiction. Therefore, $f(n)$ must not change sign for integer $n > N'$.



        So, assume WLOG that $f(n)$ is eventually positive. Note this implies for $n > N'$ and $r in mathbb{N}$, that $f(n+r) > f(n)+ rm$.



        Write
        $$sum_{n=1}^{infty} frac{1}{f(n)} = sum_{n=1}^{N'} frac{1}{f(n)} + sum_{n= N'+1}^{infty} frac{1}{f(n)}$$



        Realize that for the following series whose terms are strictly positive,
        $$sum_{n= N'+1}^{infty} frac{1}{f(n)} > sum_{n = 0}^{infty} frac{1}{f(N'+1) + nm},$$



        and the RHS obviously diverges. Contradiction.






        share|cite|improve this answer









        $endgroup$



        Assume that $sum 1/f$ converges. Then for some $N_k in mathbb{N}$,
        $$n > N_k implies |1/f(n)| < 1/k implies |f(n)| > k.$$



        $f$ is uniformly continuous. Therefore, there exists $m in mathbb{N}$ such that $|x-y| < 1/m implies |f(x)-f(y)| < 1$. By the triangle inequality, this implies $|f(n+1)-f(n)|<m$ for all $n$.



        Then if $a, a+1 > N_{lceil m/2 rceil} = N'$ are integers such that WLOG $f(a) > 0$ and $f(a+1) < 0$, then $f(a) > k$ and $f(a+1) < -k$. Then $|f(a+1)-f(a)|> 2lceil m/2 rceil > m$, contradiction. Therefore, $f(n)$ must not change sign for integer $n > N'$.



        So, assume WLOG that $f(n)$ is eventually positive. Note this implies for $n > N'$ and $r in mathbb{N}$, that $f(n+r) > f(n)+ rm$.



        Write
        $$sum_{n=1}^{infty} frac{1}{f(n)} = sum_{n=1}^{N'} frac{1}{f(n)} + sum_{n= N'+1}^{infty} frac{1}{f(n)}$$



        Realize that for the following series whose terms are strictly positive,
        $$sum_{n= N'+1}^{infty} frac{1}{f(n)} > sum_{n = 0}^{infty} frac{1}{f(N'+1) + nm},$$



        and the RHS obviously diverges. Contradiction.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 22:17









        Jeffery Opoku-MensahJeffery Opoku-Mensah

        3,536820




        3,536820






























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