For $a_n$ positive, $prod_n frac1{1+a_n}=0$ if and only if $sum_n a_n=infty$.












4












$begingroup$


This makes sense intuitively but I am not quite sure how to prove it rigorously. Any hints on how to proceed?






real-analysis sequences-and-series







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    It is helpful to note that, for $a_n$ sufficiently small, we have $frac 1{1+a_n} = 1 - a_n + a_n^2 + cdots approx 1 - a_n$
    $endgroup$
    – Omnomnomnom
    Dec 31 '18 at 20:40








  • 4




    $begingroup$
    Alternatively, I find it easier to look at the log $$ logleft[ prod_n frac 1{1 + a_n}right] = -sum_n log(1 + a_n) $$
    $endgroup$
    – Omnomnomnom
    Dec 31 '18 at 20:43










  • $begingroup$
    @Omnomnomnom Ah of course, I was forgetting that simplification with logarithm. Thanks!
    $endgroup$
    – Math1000
    Dec 31 '18 at 20:48










  • $begingroup$
    It's in the old classic Infinite Sequences And Series, by Bromwich, which I think is available cheap from Dover Publications. This book has no pre-requisite (e.g. no calculus needed). Some editions have slightly different titles. A "companion" theorem is: If $0leq a_n<1$ for all $n$ then $ prod_{n=1}^{infty}(1-a_n)=0 $ iff $ sum_{n=1}^{infty}a_n=infty.$
    $endgroup$
    – DanielWainfleet
    Dec 31 '18 at 20:57


















4












$begingroup$


This makes sense intuitively but I am not quite sure how to prove it rigorously. Any hints on how to proceed?






real-analysis sequences-and-series







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    It is helpful to note that, for $a_n$ sufficiently small, we have $frac 1{1+a_n} = 1 - a_n + a_n^2 + cdots approx 1 - a_n$
    $endgroup$
    – Omnomnomnom
    Dec 31 '18 at 20:40








  • 4




    $begingroup$
    Alternatively, I find it easier to look at the log $$ logleft[ prod_n frac 1{1 + a_n}right] = -sum_n log(1 + a_n) $$
    $endgroup$
    – Omnomnomnom
    Dec 31 '18 at 20:43










  • $begingroup$
    @Omnomnomnom Ah of course, I was forgetting that simplification with logarithm. Thanks!
    $endgroup$
    – Math1000
    Dec 31 '18 at 20:48










  • $begingroup$
    It's in the old classic Infinite Sequences And Series, by Bromwich, which I think is available cheap from Dover Publications. This book has no pre-requisite (e.g. no calculus needed). Some editions have slightly different titles. A "companion" theorem is: If $0leq a_n<1$ for all $n$ then $ prod_{n=1}^{infty}(1-a_n)=0 $ iff $ sum_{n=1}^{infty}a_n=infty.$
    $endgroup$
    – DanielWainfleet
    Dec 31 '18 at 20:57
















4












4








4


1



$begingroup$


This makes sense intuitively but I am not quite sure how to prove it rigorously. Any hints on how to proceed?






real-analysis sequences-and-series







share|cite|improve this question









$endgroup$




This makes sense intuitively but I am not quite sure how to prove it rigorously. Any hints on how to proceed?






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 31 '18 at 20:39









Math1000Math1000

19.4k31746




19.4k31746








  • 1




    $begingroup$
    It is helpful to note that, for $a_n$ sufficiently small, we have $frac 1{1+a_n} = 1 - a_n + a_n^2 + cdots approx 1 - a_n$
    $endgroup$
    – Omnomnomnom
    Dec 31 '18 at 20:40








  • 4




    $begingroup$
    Alternatively, I find it easier to look at the log $$ logleft[ prod_n frac 1{1 + a_n}right] = -sum_n log(1 + a_n) $$
    $endgroup$
    – Omnomnomnom
    Dec 31 '18 at 20:43










  • $begingroup$
    @Omnomnomnom Ah of course, I was forgetting that simplification with logarithm. Thanks!
    $endgroup$
    – Math1000
    Dec 31 '18 at 20:48










  • $begingroup$
    It's in the old classic Infinite Sequences And Series, by Bromwich, which I think is available cheap from Dover Publications. This book has no pre-requisite (e.g. no calculus needed). Some editions have slightly different titles. A "companion" theorem is: If $0leq a_n<1$ for all $n$ then $ prod_{n=1}^{infty}(1-a_n)=0 $ iff $ sum_{n=1}^{infty}a_n=infty.$
    $endgroup$
    – DanielWainfleet
    Dec 31 '18 at 20:57
















  • 1




    $begingroup$
    It is helpful to note that, for $a_n$ sufficiently small, we have $frac 1{1+a_n} = 1 - a_n + a_n^2 + cdots approx 1 - a_n$
    $endgroup$
    – Omnomnomnom
    Dec 31 '18 at 20:40








  • 4




    $begingroup$
    Alternatively, I find it easier to look at the log $$ logleft[ prod_n frac 1{1 + a_n}right] = -sum_n log(1 + a_n) $$
    $endgroup$
    – Omnomnomnom
    Dec 31 '18 at 20:43










  • $begingroup$
    @Omnomnomnom Ah of course, I was forgetting that simplification with logarithm. Thanks!
    $endgroup$
    – Math1000
    Dec 31 '18 at 20:48










  • $begingroup$
    It's in the old classic Infinite Sequences And Series, by Bromwich, which I think is available cheap from Dover Publications. This book has no pre-requisite (e.g. no calculus needed). Some editions have slightly different titles. A "companion" theorem is: If $0leq a_n<1$ for all $n$ then $ prod_{n=1}^{infty}(1-a_n)=0 $ iff $ sum_{n=1}^{infty}a_n=infty.$
    $endgroup$
    – DanielWainfleet
    Dec 31 '18 at 20:57










1




1




$begingroup$
It is helpful to note that, for $a_n$ sufficiently small, we have $frac 1{1+a_n} = 1 - a_n + a_n^2 + cdots approx 1 - a_n$
$endgroup$
– Omnomnomnom
Dec 31 '18 at 20:40






$begingroup$
It is helpful to note that, for $a_n$ sufficiently small, we have $frac 1{1+a_n} = 1 - a_n + a_n^2 + cdots approx 1 - a_n$
$endgroup$
– Omnomnomnom
Dec 31 '18 at 20:40






4




4




$begingroup$
Alternatively, I find it easier to look at the log $$ logleft[ prod_n frac 1{1 + a_n}right] = -sum_n log(1 + a_n) $$
$endgroup$
– Omnomnomnom
Dec 31 '18 at 20:43




$begingroup$
Alternatively, I find it easier to look at the log $$ logleft[ prod_n frac 1{1 + a_n}right] = -sum_n log(1 + a_n) $$
$endgroup$
– Omnomnomnom
Dec 31 '18 at 20:43












$begingroup$
@Omnomnomnom Ah of course, I was forgetting that simplification with logarithm. Thanks!
$endgroup$
– Math1000
Dec 31 '18 at 20:48




$begingroup$
@Omnomnomnom Ah of course, I was forgetting that simplification with logarithm. Thanks!
$endgroup$
– Math1000
Dec 31 '18 at 20:48












$begingroup$
It's in the old classic Infinite Sequences And Series, by Bromwich, which I think is available cheap from Dover Publications. This book has no pre-requisite (e.g. no calculus needed). Some editions have slightly different titles. A "companion" theorem is: If $0leq a_n<1$ for all $n$ then $ prod_{n=1}^{infty}(1-a_n)=0 $ iff $ sum_{n=1}^{infty}a_n=infty.$
$endgroup$
– DanielWainfleet
Dec 31 '18 at 20:57






$begingroup$
It's in the old classic Infinite Sequences And Series, by Bromwich, which I think is available cheap from Dover Publications. This book has no pre-requisite (e.g. no calculus needed). Some editions have slightly different titles. A "companion" theorem is: If $0leq a_n<1$ for all $n$ then $ prod_{n=1}^{infty}(1-a_n)=0 $ iff $ sum_{n=1}^{infty}a_n=infty.$
$endgroup$
– DanielWainfleet
Dec 31 '18 at 20:57












3 Answers
3






active

oldest

votes


















4












$begingroup$

I assume $lim_na_n=0$ otherwise it is trivial: then
$$
lnPi_nbig({1over{1+ a_n}}big)=sum -ln(1+a_n),
$$
and this is equivalent to $-sum a_n$ since
$$ln(1+a_n)simeq a_ntext{ if }lim_na_n=0.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So $-sum_nlog(1+a_n)=-infty$ would imply that the product is zero, and being finite would imply that the product is positive? Thanks.
    $endgroup$
    – Math1000
    Dec 31 '18 at 20:55



















1












$begingroup$

Let $a_ngeq 0$ for all $n.$ By induction on $n$ we have $$prod_{j=1}(1+a_j)geq 1+sum_{j=1}^na_j.$$ So if the sum diverges then so does the product.



We have $exp(a_j)geq 1+a_j, $ so $$exp (sum_{j=1}^na_j)geq prod_{j=1}^n(1+a_j).$$ So if the sum converges then so does the product.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I think you're looking at $prod_j (1+a_j)$ as opposed to $prod_j frac1{1+a_j}$ here.
    $endgroup$
    – Math1000
    Dec 31 '18 at 22:05



















0












$begingroup$

Fact 1: For any $xgeq0$, we have $ln(1+x)leq x$.



Proof: Let $f:[0,infty)rightarrowmathbb{R}$ be defined by $f(x)=x-ln(1+x)$.
Observe that $f'(x)=frac{x}{1+x}geq0,$ so $f$ is increasing. For
any $xin[0,infty)$, we have $f(x)geq f(0)=0$.



/////////////////////////////////



Fact 2: For any $xin[0,1]$, we have $ln(1+x)geqfrac{x}{2}$.



Proof: Let $g:[0,1]rightarrowmathbb{R}$ be defined by $g(x)=ln(1+x)-frac{x}{2}$.
Observe that $g'(x)=frac{1-x}{2(1+x)}geq0$, so $g$ is increasing.
For any $xin[0,1]$, we have $g(x)geq g(0)=0$.



/////////////////////////////////



Let $P_{n}=prod_{k=1}^{n}frac{1}{1+a_{k}}$. Then
begin{eqnarray*}
-ln P_{n} & = & sum_{k=1}^{n}ln(1+a_{k})\
& leq & sum_{k=1}^{n}a_{k}.
end{eqnarray*}
If $P_{n}rightarrow0$, we have $-ln P_{n}rightarrowinfty$, so
$sum_{k=1}^{infty}a_{k}=infty$.



/////////////////////////////////////////////////



For each $k$, let $b_{k}=min(a_{k},1)leq a_{k}$. Then,
begin{eqnarray*}
-ln P_{n} & = & sum_{k=1}^{n}ln(1+a_{k})\
& geq & sum_{k=1}^{n}ln(1+b_{k})\
& geq & frac{1}{2}sum_{k=1}^{n}b_{k}.
end{eqnarray*}
If $sum_{k=1}^{infty}a_{k}=infty$, we will have $sum_{k=1}^{infty}b_{k}=infty$.
(For, consider two cases. Case 1: There exists $N$ such that $a_{k}leq 1$
whenever $kgeq N$. In this case, $sum_{k=1}^{infty}b_{k}geqsum_{k=N}^{infty}b_{k}=sum_{k=N}^{infty}a_{k}=infty$.
Case 2: There does not exist $N$ such that $a_{k}leq 1$ whenever $kgeq N$.
Then, there exists a subsequence $(a_{k_{l}})_{l}$ such that $a_{k_{l}}>1$.
We have that $sum_{k=1}^{infty}b_{k}geqsum_{l=1}^{infty}b_{k_{l}}=sum_{l=1}^{infty}1=infty$.)
Therefore, $-ln P_{n}rightarrowinfty$ and hence $exp(-ln P_{n})rightarrowinfty$.
But $exp(-ln P_{n})=frac{1}{P_{n}}$. It follows that $P_{n}rightarrow0$.






share|cite|improve this answer











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    I assume $lim_na_n=0$ otherwise it is trivial: then
    $$
    lnPi_nbig({1over{1+ a_n}}big)=sum -ln(1+a_n),
    $$
    and this is equivalent to $-sum a_n$ since
    $$ln(1+a_n)simeq a_ntext{ if }lim_na_n=0.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      So $-sum_nlog(1+a_n)=-infty$ would imply that the product is zero, and being finite would imply that the product is positive? Thanks.
      $endgroup$
      – Math1000
      Dec 31 '18 at 20:55
















    4












    $begingroup$

    I assume $lim_na_n=0$ otherwise it is trivial: then
    $$
    lnPi_nbig({1over{1+ a_n}}big)=sum -ln(1+a_n),
    $$
    and this is equivalent to $-sum a_n$ since
    $$ln(1+a_n)simeq a_ntext{ if }lim_na_n=0.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      So $-sum_nlog(1+a_n)=-infty$ would imply that the product is zero, and being finite would imply that the product is positive? Thanks.
      $endgroup$
      – Math1000
      Dec 31 '18 at 20:55














    4












    4








    4





    $begingroup$

    I assume $lim_na_n=0$ otherwise it is trivial: then
    $$
    lnPi_nbig({1over{1+ a_n}}big)=sum -ln(1+a_n),
    $$
    and this is equivalent to $-sum a_n$ since
    $$ln(1+a_n)simeq a_ntext{ if }lim_na_n=0.$$






    share|cite|improve this answer











    $endgroup$



    I assume $lim_na_n=0$ otherwise it is trivial: then
    $$
    lnPi_nbig({1over{1+ a_n}}big)=sum -ln(1+a_n),
    $$
    and this is equivalent to $-sum a_n$ since
    $$ln(1+a_n)simeq a_ntext{ if }lim_na_n=0.$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 1 at 0:00









    Daniele Tampieri

    2,5772922




    2,5772922










    answered Dec 31 '18 at 20:43









    Tsemo AristideTsemo Aristide

    59.9k11446




    59.9k11446












    • $begingroup$
      So $-sum_nlog(1+a_n)=-infty$ would imply that the product is zero, and being finite would imply that the product is positive? Thanks.
      $endgroup$
      – Math1000
      Dec 31 '18 at 20:55


















    • $begingroup$
      So $-sum_nlog(1+a_n)=-infty$ would imply that the product is zero, and being finite would imply that the product is positive? Thanks.
      $endgroup$
      – Math1000
      Dec 31 '18 at 20:55
















    $begingroup$
    So $-sum_nlog(1+a_n)=-infty$ would imply that the product is zero, and being finite would imply that the product is positive? Thanks.
    $endgroup$
    – Math1000
    Dec 31 '18 at 20:55




    $begingroup$
    So $-sum_nlog(1+a_n)=-infty$ would imply that the product is zero, and being finite would imply that the product is positive? Thanks.
    $endgroup$
    – Math1000
    Dec 31 '18 at 20:55











    1












    $begingroup$

    Let $a_ngeq 0$ for all $n.$ By induction on $n$ we have $$prod_{j=1}(1+a_j)geq 1+sum_{j=1}^na_j.$$ So if the sum diverges then so does the product.



    We have $exp(a_j)geq 1+a_j, $ so $$exp (sum_{j=1}^na_j)geq prod_{j=1}^n(1+a_j).$$ So if the sum converges then so does the product.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I think you're looking at $prod_j (1+a_j)$ as opposed to $prod_j frac1{1+a_j}$ here.
      $endgroup$
      – Math1000
      Dec 31 '18 at 22:05
















    1












    $begingroup$

    Let $a_ngeq 0$ for all $n.$ By induction on $n$ we have $$prod_{j=1}(1+a_j)geq 1+sum_{j=1}^na_j.$$ So if the sum diverges then so does the product.



    We have $exp(a_j)geq 1+a_j, $ so $$exp (sum_{j=1}^na_j)geq prod_{j=1}^n(1+a_j).$$ So if the sum converges then so does the product.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I think you're looking at $prod_j (1+a_j)$ as opposed to $prod_j frac1{1+a_j}$ here.
      $endgroup$
      – Math1000
      Dec 31 '18 at 22:05














    1












    1








    1





    $begingroup$

    Let $a_ngeq 0$ for all $n.$ By induction on $n$ we have $$prod_{j=1}(1+a_j)geq 1+sum_{j=1}^na_j.$$ So if the sum diverges then so does the product.



    We have $exp(a_j)geq 1+a_j, $ so $$exp (sum_{j=1}^na_j)geq prod_{j=1}^n(1+a_j).$$ So if the sum converges then so does the product.






    share|cite|improve this answer









    $endgroup$



    Let $a_ngeq 0$ for all $n.$ By induction on $n$ we have $$prod_{j=1}(1+a_j)geq 1+sum_{j=1}^na_j.$$ So if the sum diverges then so does the product.



    We have $exp(a_j)geq 1+a_j, $ so $$exp (sum_{j=1}^na_j)geq prod_{j=1}^n(1+a_j).$$ So if the sum converges then so does the product.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 31 '18 at 21:11









    DanielWainfleetDanielWainfleet

    35.7k31648




    35.7k31648












    • $begingroup$
      I think you're looking at $prod_j (1+a_j)$ as opposed to $prod_j frac1{1+a_j}$ here.
      $endgroup$
      – Math1000
      Dec 31 '18 at 22:05


















    • $begingroup$
      I think you're looking at $prod_j (1+a_j)$ as opposed to $prod_j frac1{1+a_j}$ here.
      $endgroup$
      – Math1000
      Dec 31 '18 at 22:05
















    $begingroup$
    I think you're looking at $prod_j (1+a_j)$ as opposed to $prod_j frac1{1+a_j}$ here.
    $endgroup$
    – Math1000
    Dec 31 '18 at 22:05




    $begingroup$
    I think you're looking at $prod_j (1+a_j)$ as opposed to $prod_j frac1{1+a_j}$ here.
    $endgroup$
    – Math1000
    Dec 31 '18 at 22:05











    0












    $begingroup$

    Fact 1: For any $xgeq0$, we have $ln(1+x)leq x$.



    Proof: Let $f:[0,infty)rightarrowmathbb{R}$ be defined by $f(x)=x-ln(1+x)$.
    Observe that $f'(x)=frac{x}{1+x}geq0,$ so $f$ is increasing. For
    any $xin[0,infty)$, we have $f(x)geq f(0)=0$.



    /////////////////////////////////



    Fact 2: For any $xin[0,1]$, we have $ln(1+x)geqfrac{x}{2}$.



    Proof: Let $g:[0,1]rightarrowmathbb{R}$ be defined by $g(x)=ln(1+x)-frac{x}{2}$.
    Observe that $g'(x)=frac{1-x}{2(1+x)}geq0$, so $g$ is increasing.
    For any $xin[0,1]$, we have $g(x)geq g(0)=0$.



    /////////////////////////////////



    Let $P_{n}=prod_{k=1}^{n}frac{1}{1+a_{k}}$. Then
    begin{eqnarray*}
    -ln P_{n} & = & sum_{k=1}^{n}ln(1+a_{k})\
    & leq & sum_{k=1}^{n}a_{k}.
    end{eqnarray*}
    If $P_{n}rightarrow0$, we have $-ln P_{n}rightarrowinfty$, so
    $sum_{k=1}^{infty}a_{k}=infty$.



    /////////////////////////////////////////////////



    For each $k$, let $b_{k}=min(a_{k},1)leq a_{k}$. Then,
    begin{eqnarray*}
    -ln P_{n} & = & sum_{k=1}^{n}ln(1+a_{k})\
    & geq & sum_{k=1}^{n}ln(1+b_{k})\
    & geq & frac{1}{2}sum_{k=1}^{n}b_{k}.
    end{eqnarray*}
    If $sum_{k=1}^{infty}a_{k}=infty$, we will have $sum_{k=1}^{infty}b_{k}=infty$.
    (For, consider two cases. Case 1: There exists $N$ such that $a_{k}leq 1$
    whenever $kgeq N$. In this case, $sum_{k=1}^{infty}b_{k}geqsum_{k=N}^{infty}b_{k}=sum_{k=N}^{infty}a_{k}=infty$.
    Case 2: There does not exist $N$ such that $a_{k}leq 1$ whenever $kgeq N$.
    Then, there exists a subsequence $(a_{k_{l}})_{l}$ such that $a_{k_{l}}>1$.
    We have that $sum_{k=1}^{infty}b_{k}geqsum_{l=1}^{infty}b_{k_{l}}=sum_{l=1}^{infty}1=infty$.)
    Therefore, $-ln P_{n}rightarrowinfty$ and hence $exp(-ln P_{n})rightarrowinfty$.
    But $exp(-ln P_{n})=frac{1}{P_{n}}$. It follows that $P_{n}rightarrow0$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Fact 1: For any $xgeq0$, we have $ln(1+x)leq x$.



      Proof: Let $f:[0,infty)rightarrowmathbb{R}$ be defined by $f(x)=x-ln(1+x)$.
      Observe that $f'(x)=frac{x}{1+x}geq0,$ so $f$ is increasing. For
      any $xin[0,infty)$, we have $f(x)geq f(0)=0$.



      /////////////////////////////////



      Fact 2: For any $xin[0,1]$, we have $ln(1+x)geqfrac{x}{2}$.



      Proof: Let $g:[0,1]rightarrowmathbb{R}$ be defined by $g(x)=ln(1+x)-frac{x}{2}$.
      Observe that $g'(x)=frac{1-x}{2(1+x)}geq0$, so $g$ is increasing.
      For any $xin[0,1]$, we have $g(x)geq g(0)=0$.



      /////////////////////////////////



      Let $P_{n}=prod_{k=1}^{n}frac{1}{1+a_{k}}$. Then
      begin{eqnarray*}
      -ln P_{n} & = & sum_{k=1}^{n}ln(1+a_{k})\
      & leq & sum_{k=1}^{n}a_{k}.
      end{eqnarray*}
      If $P_{n}rightarrow0$, we have $-ln P_{n}rightarrowinfty$, so
      $sum_{k=1}^{infty}a_{k}=infty$.



      /////////////////////////////////////////////////



      For each $k$, let $b_{k}=min(a_{k},1)leq a_{k}$. Then,
      begin{eqnarray*}
      -ln P_{n} & = & sum_{k=1}^{n}ln(1+a_{k})\
      & geq & sum_{k=1}^{n}ln(1+b_{k})\
      & geq & frac{1}{2}sum_{k=1}^{n}b_{k}.
      end{eqnarray*}
      If $sum_{k=1}^{infty}a_{k}=infty$, we will have $sum_{k=1}^{infty}b_{k}=infty$.
      (For, consider two cases. Case 1: There exists $N$ such that $a_{k}leq 1$
      whenever $kgeq N$. In this case, $sum_{k=1}^{infty}b_{k}geqsum_{k=N}^{infty}b_{k}=sum_{k=N}^{infty}a_{k}=infty$.
      Case 2: There does not exist $N$ such that $a_{k}leq 1$ whenever $kgeq N$.
      Then, there exists a subsequence $(a_{k_{l}})_{l}$ such that $a_{k_{l}}>1$.
      We have that $sum_{k=1}^{infty}b_{k}geqsum_{l=1}^{infty}b_{k_{l}}=sum_{l=1}^{infty}1=infty$.)
      Therefore, $-ln P_{n}rightarrowinfty$ and hence $exp(-ln P_{n})rightarrowinfty$.
      But $exp(-ln P_{n})=frac{1}{P_{n}}$. It follows that $P_{n}rightarrow0$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Fact 1: For any $xgeq0$, we have $ln(1+x)leq x$.



        Proof: Let $f:[0,infty)rightarrowmathbb{R}$ be defined by $f(x)=x-ln(1+x)$.
        Observe that $f'(x)=frac{x}{1+x}geq0,$ so $f$ is increasing. For
        any $xin[0,infty)$, we have $f(x)geq f(0)=0$.



        /////////////////////////////////



        Fact 2: For any $xin[0,1]$, we have $ln(1+x)geqfrac{x}{2}$.



        Proof: Let $g:[0,1]rightarrowmathbb{R}$ be defined by $g(x)=ln(1+x)-frac{x}{2}$.
        Observe that $g'(x)=frac{1-x}{2(1+x)}geq0$, so $g$ is increasing.
        For any $xin[0,1]$, we have $g(x)geq g(0)=0$.



        /////////////////////////////////



        Let $P_{n}=prod_{k=1}^{n}frac{1}{1+a_{k}}$. Then
        begin{eqnarray*}
        -ln P_{n} & = & sum_{k=1}^{n}ln(1+a_{k})\
        & leq & sum_{k=1}^{n}a_{k}.
        end{eqnarray*}
        If $P_{n}rightarrow0$, we have $-ln P_{n}rightarrowinfty$, so
        $sum_{k=1}^{infty}a_{k}=infty$.



        /////////////////////////////////////////////////



        For each $k$, let $b_{k}=min(a_{k},1)leq a_{k}$. Then,
        begin{eqnarray*}
        -ln P_{n} & = & sum_{k=1}^{n}ln(1+a_{k})\
        & geq & sum_{k=1}^{n}ln(1+b_{k})\
        & geq & frac{1}{2}sum_{k=1}^{n}b_{k}.
        end{eqnarray*}
        If $sum_{k=1}^{infty}a_{k}=infty$, we will have $sum_{k=1}^{infty}b_{k}=infty$.
        (For, consider two cases. Case 1: There exists $N$ such that $a_{k}leq 1$
        whenever $kgeq N$. In this case, $sum_{k=1}^{infty}b_{k}geqsum_{k=N}^{infty}b_{k}=sum_{k=N}^{infty}a_{k}=infty$.
        Case 2: There does not exist $N$ such that $a_{k}leq 1$ whenever $kgeq N$.
        Then, there exists a subsequence $(a_{k_{l}})_{l}$ such that $a_{k_{l}}>1$.
        We have that $sum_{k=1}^{infty}b_{k}geqsum_{l=1}^{infty}b_{k_{l}}=sum_{l=1}^{infty}1=infty$.)
        Therefore, $-ln P_{n}rightarrowinfty$ and hence $exp(-ln P_{n})rightarrowinfty$.
        But $exp(-ln P_{n})=frac{1}{P_{n}}$. It follows that $P_{n}rightarrow0$.






        share|cite|improve this answer











        $endgroup$



        Fact 1: For any $xgeq0$, we have $ln(1+x)leq x$.



        Proof: Let $f:[0,infty)rightarrowmathbb{R}$ be defined by $f(x)=x-ln(1+x)$.
        Observe that $f'(x)=frac{x}{1+x}geq0,$ so $f$ is increasing. For
        any $xin[0,infty)$, we have $f(x)geq f(0)=0$.



        /////////////////////////////////



        Fact 2: For any $xin[0,1]$, we have $ln(1+x)geqfrac{x}{2}$.



        Proof: Let $g:[0,1]rightarrowmathbb{R}$ be defined by $g(x)=ln(1+x)-frac{x}{2}$.
        Observe that $g'(x)=frac{1-x}{2(1+x)}geq0$, so $g$ is increasing.
        For any $xin[0,1]$, we have $g(x)geq g(0)=0$.



        /////////////////////////////////



        Let $P_{n}=prod_{k=1}^{n}frac{1}{1+a_{k}}$. Then
        begin{eqnarray*}
        -ln P_{n} & = & sum_{k=1}^{n}ln(1+a_{k})\
        & leq & sum_{k=1}^{n}a_{k}.
        end{eqnarray*}
        If $P_{n}rightarrow0$, we have $-ln P_{n}rightarrowinfty$, so
        $sum_{k=1}^{infty}a_{k}=infty$.



        /////////////////////////////////////////////////



        For each $k$, let $b_{k}=min(a_{k},1)leq a_{k}$. Then,
        begin{eqnarray*}
        -ln P_{n} & = & sum_{k=1}^{n}ln(1+a_{k})\
        & geq & sum_{k=1}^{n}ln(1+b_{k})\
        & geq & frac{1}{2}sum_{k=1}^{n}b_{k}.
        end{eqnarray*}
        If $sum_{k=1}^{infty}a_{k}=infty$, we will have $sum_{k=1}^{infty}b_{k}=infty$.
        (For, consider two cases. Case 1: There exists $N$ such that $a_{k}leq 1$
        whenever $kgeq N$. In this case, $sum_{k=1}^{infty}b_{k}geqsum_{k=N}^{infty}b_{k}=sum_{k=N}^{infty}a_{k}=infty$.
        Case 2: There does not exist $N$ such that $a_{k}leq 1$ whenever $kgeq N$.
        Then, there exists a subsequence $(a_{k_{l}})_{l}$ such that $a_{k_{l}}>1$.
        We have that $sum_{k=1}^{infty}b_{k}geqsum_{l=1}^{infty}b_{k_{l}}=sum_{l=1}^{infty}1=infty$.)
        Therefore, $-ln P_{n}rightarrowinfty$ and hence $exp(-ln P_{n})rightarrowinfty$.
        But $exp(-ln P_{n})=frac{1}{P_{n}}$. It follows that $P_{n}rightarrow0$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 31 '18 at 21:16

























        answered Dec 31 '18 at 21:10









        Danny Pak-Keung ChanDanny Pak-Keung Chan

        2,55938




        2,55938






























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