Is there a “greatest function” that converges?












12












$begingroup$


We just hit convergence tests in calculus, and learned that $sum_{n=1}^{infty} frac{1}{n^p}$
converges for all $p gt 1$. I thought that this was sort of a "barrier" between what converges and what diverges. Specifically, that setting $a_n=frac{1}{n^{1+epsilon}}$ is sort of the "greatest function" (I'll make this precise later) for which $sum a_n$ converges.



But, I did realize that there are functions that dominate $frac{1}{n^{1+epsilon}}$ but not $frac1n$, such as $frac{1}{nlog(n)}$. Now, the sum of that specific example diverges, but it got me wondering about whether $frac{1}{n}$ is truly the "boundary". So, this leads me to two questions.



1) Is there a function $f$ that dominates $frac{1}{n^p}$ for all $p>1$, meaning: $$lim_{xtoinfty} frac{f(x)}{frac{1}{x^p}}=infty$$
Such that:
$$sum_{n=1}^infty f(n)$$ converges?



2) If so, up to a constant is there a function $g$ such that $sum_{n=1}^infty g(n)$ converges, such that $g$ dominates $f$ for all other functions $f$ such that $sum_{n=1}^infty f(n)$ converges?



I'm just a freshman in high school so I apologize if this is a stupid question.






calculus sequences-and-series convergence







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I tried to format the question such that it excludes all the "trivial" answers, i.e. "no, because you can just take this function and multiply by any constant to get another function that dominates it and still converges". The constant thing is just excluded by the fact that the limit has to diverge, rather than be equal to some constant, but something similar could still possibly happen. I tried to include a sort of "parameter" on $f$ to patch this, but I couldn't get it formal. But the sort of "philosophy" of the question is about the form of the function, rather than a "cop-out" like that.
    $endgroup$
    – Electric Moccasins
    3 hours ago












  • $begingroup$
    How about $f(x) = frac{1}{xln^2(x)}$ for your first question
    $endgroup$
    – Jakobian
    3 hours ago








  • 2




    $begingroup$
    It's a good question; part of a good answer will surely be to identify the most fruitful way to make the vague intuition behind it more precise.
    $endgroup$
    – Henning Makholm
    3 hours ago










  • $begingroup$
    From math.stackexchange.com/a/452074/42969: Let $sum_{n=1}^{infty} c_n$ be any convergent series with positive terms. Then, there exists a convergent series $sum_{n=1}^{infty} C_n$ with much bigger terms in the sense that $lim_{nrightarrowinfty} C_n/c_n = infty$.
    $endgroup$
    – Martin R
    3 hours ago
















12












$begingroup$


We just hit convergence tests in calculus, and learned that $sum_{n=1}^{infty} frac{1}{n^p}$
converges for all $p gt 1$. I thought that this was sort of a "barrier" between what converges and what diverges. Specifically, that setting $a_n=frac{1}{n^{1+epsilon}}$ is sort of the "greatest function" (I'll make this precise later) for which $sum a_n$ converges.



But, I did realize that there are functions that dominate $frac{1}{n^{1+epsilon}}$ but not $frac1n$, such as $frac{1}{nlog(n)}$. Now, the sum of that specific example diverges, but it got me wondering about whether $frac{1}{n}$ is truly the "boundary". So, this leads me to two questions.



1) Is there a function $f$ that dominates $frac{1}{n^p}$ for all $p>1$, meaning: $$lim_{xtoinfty} frac{f(x)}{frac{1}{x^p}}=infty$$
Such that:
$$sum_{n=1}^infty f(n)$$ converges?



2) If so, up to a constant is there a function $g$ such that $sum_{n=1}^infty g(n)$ converges, such that $g$ dominates $f$ for all other functions $f$ such that $sum_{n=1}^infty f(n)$ converges?



I'm just a freshman in high school so I apologize if this is a stupid question.






calculus sequences-and-series convergence







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I tried to format the question such that it excludes all the "trivial" answers, i.e. "no, because you can just take this function and multiply by any constant to get another function that dominates it and still converges". The constant thing is just excluded by the fact that the limit has to diverge, rather than be equal to some constant, but something similar could still possibly happen. I tried to include a sort of "parameter" on $f$ to patch this, but I couldn't get it formal. But the sort of "philosophy" of the question is about the form of the function, rather than a "cop-out" like that.
    $endgroup$
    – Electric Moccasins
    3 hours ago












  • $begingroup$
    How about $f(x) = frac{1}{xln^2(x)}$ for your first question
    $endgroup$
    – Jakobian
    3 hours ago








  • 2




    $begingroup$
    It's a good question; part of a good answer will surely be to identify the most fruitful way to make the vague intuition behind it more precise.
    $endgroup$
    – Henning Makholm
    3 hours ago










  • $begingroup$
    From math.stackexchange.com/a/452074/42969: Let $sum_{n=1}^{infty} c_n$ be any convergent series with positive terms. Then, there exists a convergent series $sum_{n=1}^{infty} C_n$ with much bigger terms in the sense that $lim_{nrightarrowinfty} C_n/c_n = infty$.
    $endgroup$
    – Martin R
    3 hours ago














12












12








12


1



$begingroup$


We just hit convergence tests in calculus, and learned that $sum_{n=1}^{infty} frac{1}{n^p}$
converges for all $p gt 1$. I thought that this was sort of a "barrier" between what converges and what diverges. Specifically, that setting $a_n=frac{1}{n^{1+epsilon}}$ is sort of the "greatest function" (I'll make this precise later) for which $sum a_n$ converges.



But, I did realize that there are functions that dominate $frac{1}{n^{1+epsilon}}$ but not $frac1n$, such as $frac{1}{nlog(n)}$. Now, the sum of that specific example diverges, but it got me wondering about whether $frac{1}{n}$ is truly the "boundary". So, this leads me to two questions.



1) Is there a function $f$ that dominates $frac{1}{n^p}$ for all $p>1$, meaning: $$lim_{xtoinfty} frac{f(x)}{frac{1}{x^p}}=infty$$
Such that:
$$sum_{n=1}^infty f(n)$$ converges?



2) If so, up to a constant is there a function $g$ such that $sum_{n=1}^infty g(n)$ converges, such that $g$ dominates $f$ for all other functions $f$ such that $sum_{n=1}^infty f(n)$ converges?



I'm just a freshman in high school so I apologize if this is a stupid question.






calculus sequences-and-series convergence







share|cite|improve this question









$endgroup$




We just hit convergence tests in calculus, and learned that $sum_{n=1}^{infty} frac{1}{n^p}$
converges for all $p gt 1$. I thought that this was sort of a "barrier" between what converges and what diverges. Specifically, that setting $a_n=frac{1}{n^{1+epsilon}}$ is sort of the "greatest function" (I'll make this precise later) for which $sum a_n$ converges.



But, I did realize that there are functions that dominate $frac{1}{n^{1+epsilon}}$ but not $frac1n$, such as $frac{1}{nlog(n)}$. Now, the sum of that specific example diverges, but it got me wondering about whether $frac{1}{n}$ is truly the "boundary". So, this leads me to two questions.



1) Is there a function $f$ that dominates $frac{1}{n^p}$ for all $p>1$, meaning: $$lim_{xtoinfty} frac{f(x)}{frac{1}{x^p}}=infty$$
Such that:
$$sum_{n=1}^infty f(n)$$ converges?



2) If so, up to a constant is there a function $g$ such that $sum_{n=1}^infty g(n)$ converges, such that $g$ dominates $f$ for all other functions $f$ such that $sum_{n=1}^infty f(n)$ converges?



I'm just a freshman in high school so I apologize if this is a stupid question.






calculus sequences-and-series convergence




calculus sequences-and-series convergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 3 hours ago









Electric MoccasinsElectric Moccasins

863




863








  • 1




    $begingroup$
    I tried to format the question such that it excludes all the "trivial" answers, i.e. "no, because you can just take this function and multiply by any constant to get another function that dominates it and still converges". The constant thing is just excluded by the fact that the limit has to diverge, rather than be equal to some constant, but something similar could still possibly happen. I tried to include a sort of "parameter" on $f$ to patch this, but I couldn't get it formal. But the sort of "philosophy" of the question is about the form of the function, rather than a "cop-out" like that.
    $endgroup$
    – Electric Moccasins
    3 hours ago












  • $begingroup$
    How about $f(x) = frac{1}{xln^2(x)}$ for your first question
    $endgroup$
    – Jakobian
    3 hours ago








  • 2




    $begingroup$
    It's a good question; part of a good answer will surely be to identify the most fruitful way to make the vague intuition behind it more precise.
    $endgroup$
    – Henning Makholm
    3 hours ago










  • $begingroup$
    From math.stackexchange.com/a/452074/42969: Let $sum_{n=1}^{infty} c_n$ be any convergent series with positive terms. Then, there exists a convergent series $sum_{n=1}^{infty} C_n$ with much bigger terms in the sense that $lim_{nrightarrowinfty} C_n/c_n = infty$.
    $endgroup$
    – Martin R
    3 hours ago














  • 1




    $begingroup$
    I tried to format the question such that it excludes all the "trivial" answers, i.e. "no, because you can just take this function and multiply by any constant to get another function that dominates it and still converges". The constant thing is just excluded by the fact that the limit has to diverge, rather than be equal to some constant, but something similar could still possibly happen. I tried to include a sort of "parameter" on $f$ to patch this, but I couldn't get it formal. But the sort of "philosophy" of the question is about the form of the function, rather than a "cop-out" like that.
    $endgroup$
    – Electric Moccasins
    3 hours ago












  • $begingroup$
    How about $f(x) = frac{1}{xln^2(x)}$ for your first question
    $endgroup$
    – Jakobian
    3 hours ago








  • 2




    $begingroup$
    It's a good question; part of a good answer will surely be to identify the most fruitful way to make the vague intuition behind it more precise.
    $endgroup$
    – Henning Makholm
    3 hours ago










  • $begingroup$
    From math.stackexchange.com/a/452074/42969: Let $sum_{n=1}^{infty} c_n$ be any convergent series with positive terms. Then, there exists a convergent series $sum_{n=1}^{infty} C_n$ with much bigger terms in the sense that $lim_{nrightarrowinfty} C_n/c_n = infty$.
    $endgroup$
    – Martin R
    3 hours ago








1




1




$begingroup$
I tried to format the question such that it excludes all the "trivial" answers, i.e. "no, because you can just take this function and multiply by any constant to get another function that dominates it and still converges". The constant thing is just excluded by the fact that the limit has to diverge, rather than be equal to some constant, but something similar could still possibly happen. I tried to include a sort of "parameter" on $f$ to patch this, but I couldn't get it formal. But the sort of "philosophy" of the question is about the form of the function, rather than a "cop-out" like that.
$endgroup$
– Electric Moccasins
3 hours ago






$begingroup$
I tried to format the question such that it excludes all the "trivial" answers, i.e. "no, because you can just take this function and multiply by any constant to get another function that dominates it and still converges". The constant thing is just excluded by the fact that the limit has to diverge, rather than be equal to some constant, but something similar could still possibly happen. I tried to include a sort of "parameter" on $f$ to patch this, but I couldn't get it formal. But the sort of "philosophy" of the question is about the form of the function, rather than a "cop-out" like that.
$endgroup$
– Electric Moccasins
3 hours ago














$begingroup$
How about $f(x) = frac{1}{xln^2(x)}$ for your first question
$endgroup$
– Jakobian
3 hours ago






$begingroup$
How about $f(x) = frac{1}{xln^2(x)}$ for your first question
$endgroup$
– Jakobian
3 hours ago






2




2




$begingroup$
It's a good question; part of a good answer will surely be to identify the most fruitful way to make the vague intuition behind it more precise.
$endgroup$
– Henning Makholm
3 hours ago




$begingroup$
It's a good question; part of a good answer will surely be to identify the most fruitful way to make the vague intuition behind it more precise.
$endgroup$
– Henning Makholm
3 hours ago












$begingroup$
From math.stackexchange.com/a/452074/42969: Let $sum_{n=1}^{infty} c_n$ be any convergent series with positive terms. Then, there exists a convergent series $sum_{n=1}^{infty} C_n$ with much bigger terms in the sense that $lim_{nrightarrowinfty} C_n/c_n = infty$.
$endgroup$
– Martin R
3 hours ago




$begingroup$
From math.stackexchange.com/a/452074/42969: Let $sum_{n=1}^{infty} c_n$ be any convergent series with positive terms. Then, there exists a convergent series $sum_{n=1}^{infty} C_n$ with much bigger terms in the sense that $lim_{nrightarrowinfty} C_n/c_n = infty$.
$endgroup$
– Martin R
3 hours ago










2 Answers
2






active

oldest

votes


















5












$begingroup$

1) Yes, $f(n)=frac{1}{n(ln{n})^2}$.



2) No. Assume $f geq 0$ and $sum_{n geq 1}{f(n)} < infty$.



Then there exists an increasing sequence $N_n$ and some constant $C > 0$ such that $sum_{N_n+1}^{N_{n+1}}{f(k)} leq C2^{-n}$.



Then set $g(n)=(p+1)f(n)$, where $N_p < n leq N_{p+1}$.



Then $sum_{$sum_{N_n+1}^{N_{n+1}}{g(k)} leq C(n+1)2^{-n}$, thus $sum_{n geq 1}{g(n)}$ is finite and $g(n) >> f(n)$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    1) $$f(n) = frac{1}{n log(n)^2}$$



    2) No. Given any $g > 0$ such that $sum_n g(n)$ converges, there is an increasing sequence $M_k$ such that $$sum_{n ge M_k} g(n) < 2^{-k}$$

    Then
    $ sum_n g(n) h(n)$ converges, where $h(n) = k$ for $M_k le n < M_{k+1}$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
      $endgroup$
      – orlp
      3 hours ago













    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    1) Yes, $f(n)=frac{1}{n(ln{n})^2}$.



    2) No. Assume $f geq 0$ and $sum_{n geq 1}{f(n)} < infty$.



    Then there exists an increasing sequence $N_n$ and some constant $C > 0$ such that $sum_{N_n+1}^{N_{n+1}}{f(k)} leq C2^{-n}$.



    Then set $g(n)=(p+1)f(n)$, where $N_p < n leq N_{p+1}$.



    Then $sum_{$sum_{N_n+1}^{N_{n+1}}{g(k)} leq C(n+1)2^{-n}$, thus $sum_{n geq 1}{g(n)}$ is finite and $g(n) >> f(n)$.






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      1) Yes, $f(n)=frac{1}{n(ln{n})^2}$.



      2) No. Assume $f geq 0$ and $sum_{n geq 1}{f(n)} < infty$.



      Then there exists an increasing sequence $N_n$ and some constant $C > 0$ such that $sum_{N_n+1}^{N_{n+1}}{f(k)} leq C2^{-n}$.



      Then set $g(n)=(p+1)f(n)$, where $N_p < n leq N_{p+1}$.



      Then $sum_{$sum_{N_n+1}^{N_{n+1}}{g(k)} leq C(n+1)2^{-n}$, thus $sum_{n geq 1}{g(n)}$ is finite and $g(n) >> f(n)$.






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        1) Yes, $f(n)=frac{1}{n(ln{n})^2}$.



        2) No. Assume $f geq 0$ and $sum_{n geq 1}{f(n)} < infty$.



        Then there exists an increasing sequence $N_n$ and some constant $C > 0$ such that $sum_{N_n+1}^{N_{n+1}}{f(k)} leq C2^{-n}$.



        Then set $g(n)=(p+1)f(n)$, where $N_p < n leq N_{p+1}$.



        Then $sum_{$sum_{N_n+1}^{N_{n+1}}{g(k)} leq C(n+1)2^{-n}$, thus $sum_{n geq 1}{g(n)}$ is finite and $g(n) >> f(n)$.






        share|cite|improve this answer









        $endgroup$



        1) Yes, $f(n)=frac{1}{n(ln{n})^2}$.



        2) No. Assume $f geq 0$ and $sum_{n geq 1}{f(n)} < infty$.



        Then there exists an increasing sequence $N_n$ and some constant $C > 0$ such that $sum_{N_n+1}^{N_{n+1}}{f(k)} leq C2^{-n}$.



        Then set $g(n)=(p+1)f(n)$, where $N_p < n leq N_{p+1}$.



        Then $sum_{$sum_{N_n+1}^{N_{n+1}}{g(k)} leq C(n+1)2^{-n}$, thus $sum_{n geq 1}{g(n)}$ is finite and $g(n) >> f(n)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 3 hours ago









        MindlackMindlack

        3,59517




        3,59517























            2












            $begingroup$

            1) $$f(n) = frac{1}{n log(n)^2}$$



            2) No. Given any $g > 0$ such that $sum_n g(n)$ converges, there is an increasing sequence $M_k$ such that $$sum_{n ge M_k} g(n) < 2^{-k}$$

            Then
            $ sum_n g(n) h(n)$ converges, where $h(n) = k$ for $M_k le n < M_{k+1}$.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
              $endgroup$
              – orlp
              3 hours ago


















            2












            $begingroup$

            1) $$f(n) = frac{1}{n log(n)^2}$$



            2) No. Given any $g > 0$ such that $sum_n g(n)$ converges, there is an increasing sequence $M_k$ such that $$sum_{n ge M_k} g(n) < 2^{-k}$$

            Then
            $ sum_n g(n) h(n)$ converges, where $h(n) = k$ for $M_k le n < M_{k+1}$.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
              $endgroup$
              – orlp
              3 hours ago
















            2












            2








            2





            $begingroup$

            1) $$f(n) = frac{1}{n log(n)^2}$$



            2) No. Given any $g > 0$ such that $sum_n g(n)$ converges, there is an increasing sequence $M_k$ such that $$sum_{n ge M_k} g(n) < 2^{-k}$$

            Then
            $ sum_n g(n) h(n)$ converges, where $h(n) = k$ for $M_k le n < M_{k+1}$.






            share|cite|improve this answer









            $endgroup$



            1) $$f(n) = frac{1}{n log(n)^2}$$



            2) No. Given any $g > 0$ such that $sum_n g(n)$ converges, there is an increasing sequence $M_k$ such that $$sum_{n ge M_k} g(n) < 2^{-k}$$

            Then
            $ sum_n g(n) h(n)$ converges, where $h(n) = k$ for $M_k le n < M_{k+1}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 3 hours ago









            Robert IsraelRobert Israel

            322k23212465




            322k23212465








            • 1




              $begingroup$
              If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
              $endgroup$
              – orlp
              3 hours ago
















            • 1




              $begingroup$
              If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
              $endgroup$
              – orlp
              3 hours ago










            1




            1




            $begingroup$
            If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
            $endgroup$
            – orlp
            3 hours ago






            $begingroup$
            If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
            $endgroup$
            – orlp
            3 hours ago




















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            Усть-Каменогорск

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            У этого термина существуют и другие значения, см. Усть-Каменогорск. Город Усть-Каменогорск каз. Өскемен Флаг Герб 49°57′ с. ш. 82°37′ в. д. H G Я O L Страна Казахстан   Казахстан Статус областной центр Область Восточно-Казахстанская Аким Жаксылык Омар [1] История и география Основан 1720 Город с 1868 Площадь 303,15 [2] км² Высота центра 283 ± 1 м Тип климата резко-континентальный Часовой пояс UTC+6 Население Население 329 090 [3]  человек ( 2018 ) Национальности русские 53,56 % казахи 42,99 % немцы 0,82 % украинцы 0,57 % татары 0,79 % белорусы 0,14 % азербайджанцы 0,23 % корейцы 0,20 % прочие 1,08 % -1 шт [4] Этнохороним усть-каменогорцы, усть-каменогорец Цифровые идентификаторы Телефонный код +7 (7232) Почтовый индекс F0***** [5] Автомобильный код 16 (ранее F) oskemen.kz Усть-Каменогорск Усть-Каменогорск ...
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            Халкинская богословская школа

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            Халкинская богословская школа тур. Rum Ortodoks Ruhban Okulu греч. Ιερά Θεολογική Σχολή της Χάλκης Главное здание Халкинской школы Международное название Theological School of Halki Год основания 1844 Конфессия православие Церковь Константинопольская Ректор Елпидифор (Ламбриниадис) Расположение о. Хейбелиада, Турция Юридический адрес Ümit tepesi No 5, 34973 Heybeliada, Istanbul, Turkey Сайт www.greece.org  [[commons:Category:Halki seminary|Халкинская богословская школа тур. Rum Ortodoks Ruhban Okulu греч. Ιερά Θεολογική Σχολή της Χάλκης на Викискладе]] Ха́лкинская богосло́вская шко́ла (или Духовная семинария на Халки , тур. Rum Ortodoks Ruhban Okulu , греч. Ιερά Θεολογική Σχολή της Χάλκης ) — недействующее среднее специальное учебное заведение Константинопольской православной церкви, расположенное на территории Свято-Троицкого мужского монастыря на вершине холма Надежды (тур. Ümit Tepesi ) острова Хейбелиада, в Турции....
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            Высокополье (Харьковская область)

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            У этого термина существуют и другие значения, см. Высокополье. Село Высокополье укр. Високопілля Герб 49°55′07″ с. ш. 35°25′13″ в. д. H G Я O L Страна Украина   Украина Область Харьковская Район Валковский Сельский совет Высокопольский История и география Основан 1670 Прежние названия Хмелевое Площадь 55,56 км² Часовой пояс UTC+2, летом UTC+3 Население Население 1957 человек ( 2001 ) Плотность 35,220 чел./км² Цифровые идентификаторы Телефонный код +380 5753 Почтовый индекс 63020 Автомобильный код AX, КХ / 21 КОАТУУ 6321281501 Высокополье Высокополье Высокополье  Аудио, фото и видео на Викискладе Высокопо́лье (укр. Високопілля ) — село, Высокопольский сельский совет, Валковский район, Харьковская область, Украина. Код КОАТУУ — 6321281501. Население по переписи 2001 г. составляет 1957 (866/1091 м/ж) человек. Является административным центр...
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