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









$endgroup$














    Your Answer








    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3048990%2fhow-to-prove-that-sum-k-1-infty-frackn1k-eb-n1-1-cfrac2n%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    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









    $endgroup$


















      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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3048990%2fhow-to-prove-that-sum-k-1-infty-frackn1k-eb-n1-1-cfrac2n%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
            Read more

            Puebla de Zaragoza

            Image
            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
            Read more

            Musa

            Image
            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
            Read more