Tools for finding bounds on power series

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Suppose someone hands you a series where the terms are some function of x and your goal is to find some bounds for the series for a given set of x-values (I'm thinking of power series in particular). What tools do you use to find these bounds? In particular, could you tell whether a power series in x is unbounded as x goes to infinity? I lack formal training in analysis, but please do not restrain your answers because of that.




For example, consider $$f(x) = sum_n=0^infty frac(-x)^nn!$$ and for the moment do your best to forget that this is $exp(-x)$. Let's say we looked at x = 100 and looked at the first terms of the series, we'd see some terms of rather large magnitude: $$f(100) = 1-100+5000-500000/3+... $$



The series is absolutely convergent, and the magnitude of terms eventually decreases. But how would one quickly tell that these large magnitude terms - after well over 100 such terms - would so perfectly cancel to leave a number less than 1? In the language of my question at top, how could one tell that $0 leq f(x) leq 1$ for $x geq 0$?



Here's my attempt to show this particular series is between 0 and 1 for $xgeq0$:
Consider $$g(x) = sum_m=0^infty frac(x)^mm!$$
Note that $g(x)$ is unbounded as x grows larger, and in particular that $g(x)geq 1$ for $xgeq 0$.
Noting $$g(x) f(x) = sum_n=0^infty sum_m=0^infty frac(-1)^n x^n+mn!m!,$$
we see that on changing indices to $q=n+m$ we have
$$g(x) f(x) = sum_q=0^infty sum_n=0^q frac(-1)^n x^qn!(q-n)!.$$ Thus
$$g(x) f(x) = sum_q=0^infty fracx^qq! sum_n=0^q (-1)^n binomqn,$$ which simplifies to
$$g(x) f(x) = sum_q=0^infty fracx^qq! delta_q 0 = 1.$$



Thus $$f(x) = frac1g(x),$$ and since $g(x)geq 1$ for $xgeq 0$, we see that $$0 leq f(x) leq 1 $$ for $xgeq 0$.



This attempt to put bounds on f(x) for $x in [0,infty) $ is to demonstrate my thought process on putting bounds on power series (the technique I used rested solely on foreknowledge), and I look forward to reading about the tools you use to put bounds on more general power series.







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    Suppose someone hands you a series where the terms are some function of x and your goal is to find some bounds for the series for a given set of x-values (I'm thinking of power series in particular). What tools do you use to find these bounds? In particular, could you tell whether a power series in x is unbounded as x goes to infinity? I lack formal training in analysis, but please do not restrain your answers because of that.




    For example, consider $$f(x) = sum_n=0^infty frac(-x)^nn!$$ and for the moment do your best to forget that this is $exp(-x)$. Let's say we looked at x = 100 and looked at the first terms of the series, we'd see some terms of rather large magnitude: $$f(100) = 1-100+5000-500000/3+... $$



    The series is absolutely convergent, and the magnitude of terms eventually decreases. But how would one quickly tell that these large magnitude terms - after well over 100 such terms - would so perfectly cancel to leave a number less than 1? In the language of my question at top, how could one tell that $0 leq f(x) leq 1$ for $x geq 0$?



    Here's my attempt to show this particular series is between 0 and 1 for $xgeq0$:
    Consider $$g(x) = sum_m=0^infty frac(x)^mm!$$
    Note that $g(x)$ is unbounded as x grows larger, and in particular that $g(x)geq 1$ for $xgeq 0$.
    Noting $$g(x) f(x) = sum_n=0^infty sum_m=0^infty frac(-1)^n x^n+mn!m!,$$
    we see that on changing indices to $q=n+m$ we have
    $$g(x) f(x) = sum_q=0^infty sum_n=0^q frac(-1)^n x^qn!(q-n)!.$$ Thus
    $$g(x) f(x) = sum_q=0^infty fracx^qq! sum_n=0^q (-1)^n binomqn,$$ which simplifies to
    $$g(x) f(x) = sum_q=0^infty fracx^qq! delta_q 0 = 1.$$



    Thus $$f(x) = frac1g(x),$$ and since $g(x)geq 1$ for $xgeq 0$, we see that $$0 leq f(x) leq 1 $$ for $xgeq 0$.



    This attempt to put bounds on f(x) for $x in [0,infty) $ is to demonstrate my thought process on putting bounds on power series (the technique I used rested solely on foreknowledge), and I look forward to reading about the tools you use to put bounds on more general power series.







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      up vote
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      Suppose someone hands you a series where the terms are some function of x and your goal is to find some bounds for the series for a given set of x-values (I'm thinking of power series in particular). What tools do you use to find these bounds? In particular, could you tell whether a power series in x is unbounded as x goes to infinity? I lack formal training in analysis, but please do not restrain your answers because of that.




      For example, consider $$f(x) = sum_n=0^infty frac(-x)^nn!$$ and for the moment do your best to forget that this is $exp(-x)$. Let's say we looked at x = 100 and looked at the first terms of the series, we'd see some terms of rather large magnitude: $$f(100) = 1-100+5000-500000/3+... $$



      The series is absolutely convergent, and the magnitude of terms eventually decreases. But how would one quickly tell that these large magnitude terms - after well over 100 such terms - would so perfectly cancel to leave a number less than 1? In the language of my question at top, how could one tell that $0 leq f(x) leq 1$ for $x geq 0$?



      Here's my attempt to show this particular series is between 0 and 1 for $xgeq0$:
      Consider $$g(x) = sum_m=0^infty frac(x)^mm!$$
      Note that $g(x)$ is unbounded as x grows larger, and in particular that $g(x)geq 1$ for $xgeq 0$.
      Noting $$g(x) f(x) = sum_n=0^infty sum_m=0^infty frac(-1)^n x^n+mn!m!,$$
      we see that on changing indices to $q=n+m$ we have
      $$g(x) f(x) = sum_q=0^infty sum_n=0^q frac(-1)^n x^qn!(q-n)!.$$ Thus
      $$g(x) f(x) = sum_q=0^infty fracx^qq! sum_n=0^q (-1)^n binomqn,$$ which simplifies to
      $$g(x) f(x) = sum_q=0^infty fracx^qq! delta_q 0 = 1.$$



      Thus $$f(x) = frac1g(x),$$ and since $g(x)geq 1$ for $xgeq 0$, we see that $$0 leq f(x) leq 1 $$ for $xgeq 0$.



      This attempt to put bounds on f(x) for $x in [0,infty) $ is to demonstrate my thought process on putting bounds on power series (the technique I used rested solely on foreknowledge), and I look forward to reading about the tools you use to put bounds on more general power series.







      share|cite|improve this question














      Suppose someone hands you a series where the terms are some function of x and your goal is to find some bounds for the series for a given set of x-values (I'm thinking of power series in particular). What tools do you use to find these bounds? In particular, could you tell whether a power series in x is unbounded as x goes to infinity? I lack formal training in analysis, but please do not restrain your answers because of that.




      For example, consider $$f(x) = sum_n=0^infty frac(-x)^nn!$$ and for the moment do your best to forget that this is $exp(-x)$. Let's say we looked at x = 100 and looked at the first terms of the series, we'd see some terms of rather large magnitude: $$f(100) = 1-100+5000-500000/3+... $$



      The series is absolutely convergent, and the magnitude of terms eventually decreases. But how would one quickly tell that these large magnitude terms - after well over 100 such terms - would so perfectly cancel to leave a number less than 1? In the language of my question at top, how could one tell that $0 leq f(x) leq 1$ for $x geq 0$?



      Here's my attempt to show this particular series is between 0 and 1 for $xgeq0$:
      Consider $$g(x) = sum_m=0^infty frac(x)^mm!$$
      Note that $g(x)$ is unbounded as x grows larger, and in particular that $g(x)geq 1$ for $xgeq 0$.
      Noting $$g(x) f(x) = sum_n=0^infty sum_m=0^infty frac(-1)^n x^n+mn!m!,$$
      we see that on changing indices to $q=n+m$ we have
      $$g(x) f(x) = sum_q=0^infty sum_n=0^q frac(-1)^n x^qn!(q-n)!.$$ Thus
      $$g(x) f(x) = sum_q=0^infty fracx^qq! sum_n=0^q (-1)^n binomqn,$$ which simplifies to
      $$g(x) f(x) = sum_q=0^infty fracx^qq! delta_q 0 = 1.$$



      Thus $$f(x) = frac1g(x),$$ and since $g(x)geq 1$ for $xgeq 0$, we see that $$0 leq f(x) leq 1 $$ for $xgeq 0$.



      This attempt to put bounds on f(x) for $x in [0,infty) $ is to demonstrate my thought process on putting bounds on power series (the technique I used rested solely on foreknowledge), and I look forward to reading about the tools you use to put bounds on more general power series.









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      edited Aug 24 at 18:04

























      asked Aug 24 at 3:07









      user196574

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          I'll open with one of the main analytical tools used, the comparison test. This is particularly useful for answering the 'bounded or unbounded' question.



          Suppose $A = sum_i=1^inftya_n$ and $B = sum_i=1^inftyb_n$ are series. If $a_n leq b_n$ for all $n in mathbbN$ and we know that B is bounded, then A is bounded. Similarly, if we knew that A was unbounded, we would know that B must be unbounded.



          For example, if we begin with the knowledge that $sum_i=1^inftyfrac1n$ is unbounded, we can say that $sum_i=1^inftyfrac3n$ is unbounded, as each term is greater than a known unbounded series. Similarly since $sum_i=1^inftyfrac1n^2$ is bounded, we know that $sum_i=1^inftyfrac14n^2$ is bounded, as each term is smaller than that of the former series.



          It's been a long time since I've done much analysis, there are a great many tools that other posters can hopefully explain better but I hope that's an interesting idea to get you started.






          share|cite|improve this answer




















          • Thanks for the test. I'm thinking about series that may have arbitrary sign on the terms and have dependence on some parameter "x", such as in a Taylor series. For example, with alternating terms, could one show for my f(x) that f(2x)<f(x) for x>0? In that case, the magnitude of all the terms of f(2x) is larger or equal to the magnitude of all terms in f(x), but f(2x) ends up smaller. Is there any generalization of the comparison test for series with arbitrary or alternating sign on the terms - and not just for showing convergence?
            – user196574
            Aug 24 at 17:34










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          I'll open with one of the main analytical tools used, the comparison test. This is particularly useful for answering the 'bounded or unbounded' question.



          Suppose $A = sum_i=1^inftya_n$ and $B = sum_i=1^inftyb_n$ are series. If $a_n leq b_n$ for all $n in mathbbN$ and we know that B is bounded, then A is bounded. Similarly, if we knew that A was unbounded, we would know that B must be unbounded.



          For example, if we begin with the knowledge that $sum_i=1^inftyfrac1n$ is unbounded, we can say that $sum_i=1^inftyfrac3n$ is unbounded, as each term is greater than a known unbounded series. Similarly since $sum_i=1^inftyfrac1n^2$ is bounded, we know that $sum_i=1^inftyfrac14n^2$ is bounded, as each term is smaller than that of the former series.



          It's been a long time since I've done much analysis, there are a great many tools that other posters can hopefully explain better but I hope that's an interesting idea to get you started.






          share|cite|improve this answer




















          • Thanks for the test. I'm thinking about series that may have arbitrary sign on the terms and have dependence on some parameter "x", such as in a Taylor series. For example, with alternating terms, could one show for my f(x) that f(2x)<f(x) for x>0? In that case, the magnitude of all the terms of f(2x) is larger or equal to the magnitude of all terms in f(x), but f(2x) ends up smaller. Is there any generalization of the comparison test for series with arbitrary or alternating sign on the terms - and not just for showing convergence?
            – user196574
            Aug 24 at 17:34














          up vote
          1
          down vote













          I'll open with one of the main analytical tools used, the comparison test. This is particularly useful for answering the 'bounded or unbounded' question.



          Suppose $A = sum_i=1^inftya_n$ and $B = sum_i=1^inftyb_n$ are series. If $a_n leq b_n$ for all $n in mathbbN$ and we know that B is bounded, then A is bounded. Similarly, if we knew that A was unbounded, we would know that B must be unbounded.



          For example, if we begin with the knowledge that $sum_i=1^inftyfrac1n$ is unbounded, we can say that $sum_i=1^inftyfrac3n$ is unbounded, as each term is greater than a known unbounded series. Similarly since $sum_i=1^inftyfrac1n^2$ is bounded, we know that $sum_i=1^inftyfrac14n^2$ is bounded, as each term is smaller than that of the former series.



          It's been a long time since I've done much analysis, there are a great many tools that other posters can hopefully explain better but I hope that's an interesting idea to get you started.






          share|cite|improve this answer




















          • Thanks for the test. I'm thinking about series that may have arbitrary sign on the terms and have dependence on some parameter "x", such as in a Taylor series. For example, with alternating terms, could one show for my f(x) that f(2x)<f(x) for x>0? In that case, the magnitude of all the terms of f(2x) is larger or equal to the magnitude of all terms in f(x), but f(2x) ends up smaller. Is there any generalization of the comparison test for series with arbitrary or alternating sign on the terms - and not just for showing convergence?
            – user196574
            Aug 24 at 17:34












          up vote
          1
          down vote










          up vote
          1
          down vote









          I'll open with one of the main analytical tools used, the comparison test. This is particularly useful for answering the 'bounded or unbounded' question.



          Suppose $A = sum_i=1^inftya_n$ and $B = sum_i=1^inftyb_n$ are series. If $a_n leq b_n$ for all $n in mathbbN$ and we know that B is bounded, then A is bounded. Similarly, if we knew that A was unbounded, we would know that B must be unbounded.



          For example, if we begin with the knowledge that $sum_i=1^inftyfrac1n$ is unbounded, we can say that $sum_i=1^inftyfrac3n$ is unbounded, as each term is greater than a known unbounded series. Similarly since $sum_i=1^inftyfrac1n^2$ is bounded, we know that $sum_i=1^inftyfrac14n^2$ is bounded, as each term is smaller than that of the former series.



          It's been a long time since I've done much analysis, there are a great many tools that other posters can hopefully explain better but I hope that's an interesting idea to get you started.






          share|cite|improve this answer












          I'll open with one of the main analytical tools used, the comparison test. This is particularly useful for answering the 'bounded or unbounded' question.



          Suppose $A = sum_i=1^inftya_n$ and $B = sum_i=1^inftyb_n$ are series. If $a_n leq b_n$ for all $n in mathbbN$ and we know that B is bounded, then A is bounded. Similarly, if we knew that A was unbounded, we would know that B must be unbounded.



          For example, if we begin with the knowledge that $sum_i=1^inftyfrac1n$ is unbounded, we can say that $sum_i=1^inftyfrac3n$ is unbounded, as each term is greater than a known unbounded series. Similarly since $sum_i=1^inftyfrac1n^2$ is bounded, we know that $sum_i=1^inftyfrac14n^2$ is bounded, as each term is smaller than that of the former series.



          It's been a long time since I've done much analysis, there are a great many tools that other posters can hopefully explain better but I hope that's an interesting idea to get you started.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 24 at 12:40









          Gus314

          411




          411











          • Thanks for the test. I'm thinking about series that may have arbitrary sign on the terms and have dependence on some parameter "x", such as in a Taylor series. For example, with alternating terms, could one show for my f(x) that f(2x)<f(x) for x>0? In that case, the magnitude of all the terms of f(2x) is larger or equal to the magnitude of all terms in f(x), but f(2x) ends up smaller. Is there any generalization of the comparison test for series with arbitrary or alternating sign on the terms - and not just for showing convergence?
            – user196574
            Aug 24 at 17:34
















          • Thanks for the test. I'm thinking about series that may have arbitrary sign on the terms and have dependence on some parameter "x", such as in a Taylor series. For example, with alternating terms, could one show for my f(x) that f(2x)<f(x) for x>0? In that case, the magnitude of all the terms of f(2x) is larger or equal to the magnitude of all terms in f(x), but f(2x) ends up smaller. Is there any generalization of the comparison test for series with arbitrary or alternating sign on the terms - and not just for showing convergence?
            – user196574
            Aug 24 at 17:34















          Thanks for the test. I'm thinking about series that may have arbitrary sign on the terms and have dependence on some parameter "x", such as in a Taylor series. For example, with alternating terms, could one show for my f(x) that f(2x)<f(x) for x>0? In that case, the magnitude of all the terms of f(2x) is larger or equal to the magnitude of all terms in f(x), but f(2x) ends up smaller. Is there any generalization of the comparison test for series with arbitrary or alternating sign on the terms - and not just for showing convergence?
          – user196574
          Aug 24 at 17:34




          Thanks for the test. I'm thinking about series that may have arbitrary sign on the terms and have dependence on some parameter "x", such as in a Taylor series. For example, with alternating terms, could one show for my f(x) that f(2x)<f(x) for x>0? In that case, the magnitude of all the terms of f(2x) is larger or equal to the magnitude of all terms in f(x), but f(2x) ends up smaller. Is there any generalization of the comparison test for series with arbitrary or alternating sign on the terms - and not just for showing convergence?
          – user196574
          Aug 24 at 17:34

















           

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