How to prove that...












9












$begingroup$


Through some calculation, it can be shown that
$$e = 1+cfrac{1+cfrac{1+cfrac{1+cfrac{vdots}{4}}{3}}{2}}{1}tag{1}$$
$$2e = 1+cfrac{2+cfrac{3+cfrac{4+cfrac{vdots}{4}}{3}}{2}}{1}tag{2}$$
$$5e = 1+cfrac{2^2+cfrac{3^2+cfrac{4^2+cfrac{vdots}{4}}{3}}{2}}{1}tag{3}$$
In general, how can I show that



$$sum_{k=1}^{infty}frac{k^{n+1}}{k!}=eB_{n+1}=1+cfrac{2^n+cfrac{3^n+cfrac{4^n+cfrac{vdots}{4}}{3}}{2}}{1}$$
, where $B_n$ is the $n^{th}$ Bell number.



I saw a similar question on Brilliant.org, but I did not pay close attention to the proof, and I ended up forgetting how to prove this kind of problem.
I remember that the proof involves in simplifying the denominator and moving from top to bottom so that the final denominator is in the form of $n!$, which is the criteria for Maclaurin series.



Here is the background information Infinite Series $sumlimits_{k=1}^{infty}frac{k^n}{k!}$






sequences-and-series continued-fractions







share|cite|improve this question











$endgroup$












  • $begingroup$
    See my answer here for the first equation.
    $endgroup$
    – JavaMan
    Dec 21 '18 at 23:26






  • 1




    $begingroup$
    It might be worth noting in the question that the $B_n$ are Bell numbers.
    $endgroup$
    – Carl Schildkraut
    Dec 21 '18 at 23:27










  • $begingroup$
    The last time I saw upwards continued fractions was...last millenium, I think. Is there any real advantage to upwards instead of downwards?
    $endgroup$
    – DonAntonio
    Dec 21 '18 at 23:33










  • $begingroup$
    @CarlSchildkraut: Ok, I will note that.
    $endgroup$
    – Larry
    Dec 21 '18 at 23:53
















9












$begingroup$


Through some calculation, it can be shown that
$$e = 1+cfrac{1+cfrac{1+cfrac{1+cfrac{vdots}{4}}{3}}{2}}{1}tag{1}$$
$$2e = 1+cfrac{2+cfrac{3+cfrac{4+cfrac{vdots}{4}}{3}}{2}}{1}tag{2}$$
$$5e = 1+cfrac{2^2+cfrac{3^2+cfrac{4^2+cfrac{vdots}{4}}{3}}{2}}{1}tag{3}$$
In general, how can I show that



$$sum_{k=1}^{infty}frac{k^{n+1}}{k!}=eB_{n+1}=1+cfrac{2^n+cfrac{3^n+cfrac{4^n+cfrac{vdots}{4}}{3}}{2}}{1}$$
, where $B_n$ is the $n^{th}$ Bell number.



I saw a similar question on Brilliant.org, but I did not pay close attention to the proof, and I ended up forgetting how to prove this kind of problem.
I remember that the proof involves in simplifying the denominator and moving from top to bottom so that the final denominator is in the form of $n!$, which is the criteria for Maclaurin series.



Here is the background information Infinite Series $sumlimits_{k=1}^{infty}frac{k^n}{k!}$






sequences-and-series continued-fractions







share|cite|improve this question











$endgroup$












  • $begingroup$
    See my answer here for the first equation.
    $endgroup$
    – JavaMan
    Dec 21 '18 at 23:26






  • 1




    $begingroup$
    It might be worth noting in the question that the $B_n$ are Bell numbers.
    $endgroup$
    – Carl Schildkraut
    Dec 21 '18 at 23:27










  • $begingroup$
    The last time I saw upwards continued fractions was...last millenium, I think. Is there any real advantage to upwards instead of downwards?
    $endgroup$
    – DonAntonio
    Dec 21 '18 at 23:33










  • $begingroup$
    @CarlSchildkraut: Ok, I will note that.
    $endgroup$
    – Larry
    Dec 21 '18 at 23:53














9












9








9


4



$begingroup$


Through some calculation, it can be shown that
$$e = 1+cfrac{1+cfrac{1+cfrac{1+cfrac{vdots}{4}}{3}}{2}}{1}tag{1}$$
$$2e = 1+cfrac{2+cfrac{3+cfrac{4+cfrac{vdots}{4}}{3}}{2}}{1}tag{2}$$
$$5e = 1+cfrac{2^2+cfrac{3^2+cfrac{4^2+cfrac{vdots}{4}}{3}}{2}}{1}tag{3}$$
In general, how can I show that



$$sum_{k=1}^{infty}frac{k^{n+1}}{k!}=eB_{n+1}=1+cfrac{2^n+cfrac{3^n+cfrac{4^n+cfrac{vdots}{4}}{3}}{2}}{1}$$
, where $B_n$ is the $n^{th}$ Bell number.



I saw a similar question on Brilliant.org, but I did not pay close attention to the proof, and I ended up forgetting how to prove this kind of problem.
I remember that the proof involves in simplifying the denominator and moving from top to bottom so that the final denominator is in the form of $n!$, which is the criteria for Maclaurin series.



Here is the background information Infinite Series $sumlimits_{k=1}^{infty}frac{k^n}{k!}$






sequences-and-series continued-fractions







share|cite|improve this question











$endgroup$




Through some calculation, it can be shown that
$$e = 1+cfrac{1+cfrac{1+cfrac{1+cfrac{vdots}{4}}{3}}{2}}{1}tag{1}$$
$$2e = 1+cfrac{2+cfrac{3+cfrac{4+cfrac{vdots}{4}}{3}}{2}}{1}tag{2}$$
$$5e = 1+cfrac{2^2+cfrac{3^2+cfrac{4^2+cfrac{vdots}{4}}{3}}{2}}{1}tag{3}$$
In general, how can I show that



$$sum_{k=1}^{infty}frac{k^{n+1}}{k!}=eB_{n+1}=1+cfrac{2^n+cfrac{3^n+cfrac{4^n+cfrac{vdots}{4}}{3}}{2}}{1}$$
, where $B_n$ is the $n^{th}$ Bell number.



I saw a similar question on Brilliant.org, but I did not pay close attention to the proof, and I ended up forgetting how to prove this kind of problem.
I remember that the proof involves in simplifying the denominator and moving from top to bottom so that the final denominator is in the form of $n!$, which is the criteria for Maclaurin series.



Here is the background information Infinite Series $sumlimits_{k=1}^{infty}frac{k^n}{k!}$






sequences-and-series continued-fractions




sequences-and-series continued-fractions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 2:37







Larry

















asked Dec 21 '18 at 23:18









LarryLarry

2,55031131




2,55031131












  • $begingroup$
    See my answer here for the first equation.
    $endgroup$
    – JavaMan
    Dec 21 '18 at 23:26






  • 1




    $begingroup$
    It might be worth noting in the question that the $B_n$ are Bell numbers.
    $endgroup$
    – Carl Schildkraut
    Dec 21 '18 at 23:27










  • $begingroup$
    The last time I saw upwards continued fractions was...last millenium, I think. Is there any real advantage to upwards instead of downwards?
    $endgroup$
    – DonAntonio
    Dec 21 '18 at 23:33










  • $begingroup$
    @CarlSchildkraut: Ok, I will note that.
    $endgroup$
    – Larry
    Dec 21 '18 at 23:53


















  • $begingroup$
    See my answer here for the first equation.
    $endgroup$
    – JavaMan
    Dec 21 '18 at 23:26






  • 1




    $begingroup$
    It might be worth noting in the question that the $B_n$ are Bell numbers.
    $endgroup$
    – Carl Schildkraut
    Dec 21 '18 at 23:27










  • $begingroup$
    The last time I saw upwards continued fractions was...last millenium, I think. Is there any real advantage to upwards instead of downwards?
    $endgroup$
    – DonAntonio
    Dec 21 '18 at 23:33










  • $begingroup$
    @CarlSchildkraut: Ok, I will note that.
    $endgroup$
    – Larry
    Dec 21 '18 at 23:53
















$begingroup$
See my answer here for the first equation.
$endgroup$
– JavaMan
Dec 21 '18 at 23:26




$begingroup$
See my answer here for the first equation.
$endgroup$
– JavaMan
Dec 21 '18 at 23:26




1




1




$begingroup$
It might be worth noting in the question that the $B_n$ are Bell numbers.
$endgroup$
– Carl Schildkraut
Dec 21 '18 at 23:27




$begingroup$
It might be worth noting in the question that the $B_n$ are Bell numbers.
$endgroup$
– Carl Schildkraut
Dec 21 '18 at 23:27












$begingroup$
The last time I saw upwards continued fractions was...last millenium, I think. Is there any real advantage to upwards instead of downwards?
$endgroup$
– DonAntonio
Dec 21 '18 at 23:33




$begingroup$
The last time I saw upwards continued fractions was...last millenium, I think. Is there any real advantage to upwards instead of downwards?
$endgroup$
– DonAntonio
Dec 21 '18 at 23:33












$begingroup$
@CarlSchildkraut: Ok, I will note that.
$endgroup$
– Larry
Dec 21 '18 at 23:53




$begingroup$
@CarlSchildkraut: Ok, I will note that.
$endgroup$
– Larry
Dec 21 '18 at 23:53










1 Answer
1






active

oldest

votes


















3












$begingroup$

After some work, I recalled the method from Brilliant.org. This is not a rigorous proof, but it offers some intuitive sense.



We will first prove the first equation. Note that
$$begin{align}
1+cfrac{1+cfrac{1+cfrac{1+cfrac{1+cfrac{1+vdots}{5}}{4}}{3}}{2}}{1}&=1+cfrac{1+cfrac{1+cfrac{1+cfrac{1+cfrac{1}{5}+cfrac{vdots}{5}}{4}}{3}}{2}}{1}\
&=1+cfrac{1+cfrac{1+cfrac{1+cfrac{1}{4}+cfrac{1}{4cdot5}+cfrac{vdots}{4cdot5}}{3}}{2}}{1}\
&=1+cfrac{1+cfrac{1+cfrac{1}{3}+cfrac{1}{3cdot4}+cfrac{1}{3cdot4cdot5}+cfrac{vdots}{3cdot4cdot5}}{2}}{1}\
&=1+cfrac{1+cfrac{1}{2}+cfrac{1}{2cdot3}+cfrac{1}{2cdot3cdot4}+cfrac{1}{2cdot3cdot4cdot5}+cfrac{vdots}{2cdot3cdot4cdot5}}{1}\
&=1+cfrac{1}{1!}+cfrac{1}{2!}+cfrac{1}{3!}+cfrac{1}{4!}+cfrac{1}{5!}+cdots\
&=color{red}etag{1}
end{align}$$
We will then proceed to the second equation.
$$begin{align}
1+cfrac{2+cfrac{3+cfrac{4+cfrac{5+cfrac{6+vdots}{5}}{4}}{3}}{2}}{1}&=1+cfrac{2+cfrac{3+cfrac{4+cfrac{5+cfrac{6}{5}+cfrac{vdots}{5}}{4}}{3}}{2}}{1}\
&=1+cfrac{2+cfrac{3+cfrac{4+cfrac{5}{4}+cfrac{6}{4cdot5}+cfrac{vdots}{4cdot5}}{3}}{2}}{1}\
&=1+cfrac{2+cfrac{3+cfrac{4}{3}+cfrac{5}{3cdot4}+cfrac{6}{3cdot4cdot5}+cfrac{vdots}{3cdot4cdot5}}{2}}{1}\
&=1+cfrac{2+cfrac{3}{2}+cfrac{4}{2cdot3}+cfrac{5}{2cdot3cdot4}+cfrac{6}{2cdot3cdot4cdot5}+cfrac{vdots}{2cdot3cdot4cdot5}}{1}\
&=1+cfrac{2}{1!}+cfrac{3}{2!}+cfrac{4}{3!}+cfrac{5}{4!}+cfrac{6}{5!}+cdots\
&=sum_{n=1}^{infty}cfrac{n^2}{n!}\
&=color{red}{2e}tag{2}
end{align}$$
Then, using the same logic, we have
$$sum_{k=1}^{infty}frac{k^{n+1}}{k!}=eB_{n+1}=1+cfrac{2^n+cfrac{3^n+cfrac{4^n+cfrac{vdots}{4}}{3}}{2}}{1}$$






share|cite|improve this answer









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    1 Answer
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    active

    oldest

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    3












    $begingroup$

    After some work, I recalled the method from Brilliant.org. This is not a rigorous proof, but it offers some intuitive sense.



    We will first prove the first equation. Note that
    $$begin{align}
    1+cfrac{1+cfrac{1+cfrac{1+cfrac{1+cfrac{1+vdots}{5}}{4}}{3}}{2}}{1}&=1+cfrac{1+cfrac{1+cfrac{1+cfrac{1+cfrac{1}{5}+cfrac{vdots}{5}}{4}}{3}}{2}}{1}\
    &=1+cfrac{1+cfrac{1+cfrac{1+cfrac{1}{4}+cfrac{1}{4cdot5}+cfrac{vdots}{4cdot5}}{3}}{2}}{1}\
    &=1+cfrac{1+cfrac{1+cfrac{1}{3}+cfrac{1}{3cdot4}+cfrac{1}{3cdot4cdot5}+cfrac{vdots}{3cdot4cdot5}}{2}}{1}\
    &=1+cfrac{1+cfrac{1}{2}+cfrac{1}{2cdot3}+cfrac{1}{2cdot3cdot4}+cfrac{1}{2cdot3cdot4cdot5}+cfrac{vdots}{2cdot3cdot4cdot5}}{1}\
    &=1+cfrac{1}{1!}+cfrac{1}{2!}+cfrac{1}{3!}+cfrac{1}{4!}+cfrac{1}{5!}+cdots\
    &=color{red}etag{1}
    end{align}$$
    We will then proceed to the second equation.
    $$begin{align}
    1+cfrac{2+cfrac{3+cfrac{4+cfrac{5+cfrac{6+vdots}{5}}{4}}{3}}{2}}{1}&=1+cfrac{2+cfrac{3+cfrac{4+cfrac{5+cfrac{6}{5}+cfrac{vdots}{5}}{4}}{3}}{2}}{1}\
    &=1+cfrac{2+cfrac{3+cfrac{4+cfrac{5}{4}+cfrac{6}{4cdot5}+cfrac{vdots}{4cdot5}}{3}}{2}}{1}\
    &=1+cfrac{2+cfrac{3+cfrac{4}{3}+cfrac{5}{3cdot4}+cfrac{6}{3cdot4cdot5}+cfrac{vdots}{3cdot4cdot5}}{2}}{1}\
    &=1+cfrac{2+cfrac{3}{2}+cfrac{4}{2cdot3}+cfrac{5}{2cdot3cdot4}+cfrac{6}{2cdot3cdot4cdot5}+cfrac{vdots}{2cdot3cdot4cdot5}}{1}\
    &=1+cfrac{2}{1!}+cfrac{3}{2!}+cfrac{4}{3!}+cfrac{5}{4!}+cfrac{6}{5!}+cdots\
    &=sum_{n=1}^{infty}cfrac{n^2}{n!}\
    &=color{red}{2e}tag{2}
    end{align}$$
    Then, using the same logic, we have
    $$sum_{k=1}^{infty}frac{k^{n+1}}{k!}=eB_{n+1}=1+cfrac{2^n+cfrac{3^n+cfrac{4^n+cfrac{vdots}{4}}{3}}{2}}{1}$$






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      After some work, I recalled the method from Brilliant.org. This is not a rigorous proof, but it offers some intuitive sense.



      We will first prove the first equation. Note that
      $$begin{align}
      1+cfrac{1+cfrac{1+cfrac{1+cfrac{1+cfrac{1+vdots}{5}}{4}}{3}}{2}}{1}&=1+cfrac{1+cfrac{1+cfrac{1+cfrac{1+cfrac{1}{5}+cfrac{vdots}{5}}{4}}{3}}{2}}{1}\
      &=1+cfrac{1+cfrac{1+cfrac{1+cfrac{1}{4}+cfrac{1}{4cdot5}+cfrac{vdots}{4cdot5}}{3}}{2}}{1}\
      &=1+cfrac{1+cfrac{1+cfrac{1}{3}+cfrac{1}{3cdot4}+cfrac{1}{3cdot4cdot5}+cfrac{vdots}{3cdot4cdot5}}{2}}{1}\
      &=1+cfrac{1+cfrac{1}{2}+cfrac{1}{2cdot3}+cfrac{1}{2cdot3cdot4}+cfrac{1}{2cdot3cdot4cdot5}+cfrac{vdots}{2cdot3cdot4cdot5}}{1}\
      &=1+cfrac{1}{1!}+cfrac{1}{2!}+cfrac{1}{3!}+cfrac{1}{4!}+cfrac{1}{5!}+cdots\
      &=color{red}etag{1}
      end{align}$$
      We will then proceed to the second equation.
      $$begin{align}
      1+cfrac{2+cfrac{3+cfrac{4+cfrac{5+cfrac{6+vdots}{5}}{4}}{3}}{2}}{1}&=1+cfrac{2+cfrac{3+cfrac{4+cfrac{5+cfrac{6}{5}+cfrac{vdots}{5}}{4}}{3}}{2}}{1}\
      &=1+cfrac{2+cfrac{3+cfrac{4+cfrac{5}{4}+cfrac{6}{4cdot5}+cfrac{vdots}{4cdot5}}{3}}{2}}{1}\
      &=1+cfrac{2+cfrac{3+cfrac{4}{3}+cfrac{5}{3cdot4}+cfrac{6}{3cdot4cdot5}+cfrac{vdots}{3cdot4cdot5}}{2}}{1}\
      &=1+cfrac{2+cfrac{3}{2}+cfrac{4}{2cdot3}+cfrac{5}{2cdot3cdot4}+cfrac{6}{2cdot3cdot4cdot5}+cfrac{vdots}{2cdot3cdot4cdot5}}{1}\
      &=1+cfrac{2}{1!}+cfrac{3}{2!}+cfrac{4}{3!}+cfrac{5}{4!}+cfrac{6}{5!}+cdots\
      &=sum_{n=1}^{infty}cfrac{n^2}{n!}\
      &=color{red}{2e}tag{2}
      end{align}$$
      Then, using the same logic, we have
      $$sum_{k=1}^{infty}frac{k^{n+1}}{k!}=eB_{n+1}=1+cfrac{2^n+cfrac{3^n+cfrac{4^n+cfrac{vdots}{4}}{3}}{2}}{1}$$






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        After some work, I recalled the method from Brilliant.org. This is not a rigorous proof, but it offers some intuitive sense.



        We will first prove the first equation. Note that
        $$begin{align}
        1+cfrac{1+cfrac{1+cfrac{1+cfrac{1+cfrac{1+vdots}{5}}{4}}{3}}{2}}{1}&=1+cfrac{1+cfrac{1+cfrac{1+cfrac{1+cfrac{1}{5}+cfrac{vdots}{5}}{4}}{3}}{2}}{1}\
        &=1+cfrac{1+cfrac{1+cfrac{1+cfrac{1}{4}+cfrac{1}{4cdot5}+cfrac{vdots}{4cdot5}}{3}}{2}}{1}\
        &=1+cfrac{1+cfrac{1+cfrac{1}{3}+cfrac{1}{3cdot4}+cfrac{1}{3cdot4cdot5}+cfrac{vdots}{3cdot4cdot5}}{2}}{1}\
        &=1+cfrac{1+cfrac{1}{2}+cfrac{1}{2cdot3}+cfrac{1}{2cdot3cdot4}+cfrac{1}{2cdot3cdot4cdot5}+cfrac{vdots}{2cdot3cdot4cdot5}}{1}\
        &=1+cfrac{1}{1!}+cfrac{1}{2!}+cfrac{1}{3!}+cfrac{1}{4!}+cfrac{1}{5!}+cdots\
        &=color{red}etag{1}
        end{align}$$
        We will then proceed to the second equation.
        $$begin{align}
        1+cfrac{2+cfrac{3+cfrac{4+cfrac{5+cfrac{6+vdots}{5}}{4}}{3}}{2}}{1}&=1+cfrac{2+cfrac{3+cfrac{4+cfrac{5+cfrac{6}{5}+cfrac{vdots}{5}}{4}}{3}}{2}}{1}\
        &=1+cfrac{2+cfrac{3+cfrac{4+cfrac{5}{4}+cfrac{6}{4cdot5}+cfrac{vdots}{4cdot5}}{3}}{2}}{1}\
        &=1+cfrac{2+cfrac{3+cfrac{4}{3}+cfrac{5}{3cdot4}+cfrac{6}{3cdot4cdot5}+cfrac{vdots}{3cdot4cdot5}}{2}}{1}\
        &=1+cfrac{2+cfrac{3}{2}+cfrac{4}{2cdot3}+cfrac{5}{2cdot3cdot4}+cfrac{6}{2cdot3cdot4cdot5}+cfrac{vdots}{2cdot3cdot4cdot5}}{1}\
        &=1+cfrac{2}{1!}+cfrac{3}{2!}+cfrac{4}{3!}+cfrac{5}{4!}+cfrac{6}{5!}+cdots\
        &=sum_{n=1}^{infty}cfrac{n^2}{n!}\
        &=color{red}{2e}tag{2}
        end{align}$$
        Then, using the same logic, we have
        $$sum_{k=1}^{infty}frac{k^{n+1}}{k!}=eB_{n+1}=1+cfrac{2^n+cfrac{3^n+cfrac{4^n+cfrac{vdots}{4}}{3}}{2}}{1}$$






        share|cite|improve this answer









        $endgroup$



        After some work, I recalled the method from Brilliant.org. This is not a rigorous proof, but it offers some intuitive sense.



        We will first prove the first equation. Note that
        $$begin{align}
        1+cfrac{1+cfrac{1+cfrac{1+cfrac{1+cfrac{1+vdots}{5}}{4}}{3}}{2}}{1}&=1+cfrac{1+cfrac{1+cfrac{1+cfrac{1+cfrac{1}{5}+cfrac{vdots}{5}}{4}}{3}}{2}}{1}\
        &=1+cfrac{1+cfrac{1+cfrac{1+cfrac{1}{4}+cfrac{1}{4cdot5}+cfrac{vdots}{4cdot5}}{3}}{2}}{1}\
        &=1+cfrac{1+cfrac{1+cfrac{1}{3}+cfrac{1}{3cdot4}+cfrac{1}{3cdot4cdot5}+cfrac{vdots}{3cdot4cdot5}}{2}}{1}\
        &=1+cfrac{1+cfrac{1}{2}+cfrac{1}{2cdot3}+cfrac{1}{2cdot3cdot4}+cfrac{1}{2cdot3cdot4cdot5}+cfrac{vdots}{2cdot3cdot4cdot5}}{1}\
        &=1+cfrac{1}{1!}+cfrac{1}{2!}+cfrac{1}{3!}+cfrac{1}{4!}+cfrac{1}{5!}+cdots\
        &=color{red}etag{1}
        end{align}$$
        We will then proceed to the second equation.
        $$begin{align}
        1+cfrac{2+cfrac{3+cfrac{4+cfrac{5+cfrac{6+vdots}{5}}{4}}{3}}{2}}{1}&=1+cfrac{2+cfrac{3+cfrac{4+cfrac{5+cfrac{6}{5}+cfrac{vdots}{5}}{4}}{3}}{2}}{1}\
        &=1+cfrac{2+cfrac{3+cfrac{4+cfrac{5}{4}+cfrac{6}{4cdot5}+cfrac{vdots}{4cdot5}}{3}}{2}}{1}\
        &=1+cfrac{2+cfrac{3+cfrac{4}{3}+cfrac{5}{3cdot4}+cfrac{6}{3cdot4cdot5}+cfrac{vdots}{3cdot4cdot5}}{2}}{1}\
        &=1+cfrac{2+cfrac{3}{2}+cfrac{4}{2cdot3}+cfrac{5}{2cdot3cdot4}+cfrac{6}{2cdot3cdot4cdot5}+cfrac{vdots}{2cdot3cdot4cdot5}}{1}\
        &=1+cfrac{2}{1!}+cfrac{3}{2!}+cfrac{4}{3!}+cfrac{5}{4!}+cfrac{6}{5!}+cdots\
        &=sum_{n=1}^{infty}cfrac{n^2}{n!}\
        &=color{red}{2e}tag{2}
        end{align}$$
        Then, using the same logic, we have
        $$sum_{k=1}^{infty}frac{k^{n+1}}{k!}=eB_{n+1}=1+cfrac{2^n+cfrac{3^n+cfrac{4^n+cfrac{vdots}{4}}{3}}{2}}{1}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 22 '18 at 5:40









        LarryLarry

        2,55031131




        2,55031131






























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