Alternative proof for $ sum left(frac{k-2}{k}right)^k$ is divergent.












0














In an earlier question I asked help to prove that:
$$ sum left(frac{k-2}{k}right)^k$$ is divergent, using the fact the the general term converges to $frac{1}{e^2}$ and by the limit test the series diverges, is there a more straightforward method I am missing? One that does not use $e$ necessarily.




See original post: Series diverges






real-analysis sequences-and-series alternative-proof







share|cite|improve this question


















  • 10




    I would say that proving that the general term does not converge to zero is about as simple as it gets.
    – MisterRiemann
    Nov 28 at 12:17
















0














In an earlier question I asked help to prove that:
$$ sum left(frac{k-2}{k}right)^k$$ is divergent, using the fact the the general term converges to $frac{1}{e^2}$ and by the limit test the series diverges, is there a more straightforward method I am missing? One that does not use $e$ necessarily.




See original post: Series diverges






real-analysis sequences-and-series alternative-proof







share|cite|improve this question


















  • 10




    I would say that proving that the general term does not converge to zero is about as simple as it gets.
    – MisterRiemann
    Nov 28 at 12:17














0












0








0







In an earlier question I asked help to prove that:
$$ sum left(frac{k-2}{k}right)^k$$ is divergent, using the fact the the general term converges to $frac{1}{e^2}$ and by the limit test the series diverges, is there a more straightforward method I am missing? One that does not use $e$ necessarily.




See original post: Series diverges






real-analysis sequences-and-series alternative-proof







share|cite|improve this question













In an earlier question I asked help to prove that:
$$ sum left(frac{k-2}{k}right)^k$$ is divergent, using the fact the the general term converges to $frac{1}{e^2}$ and by the limit test the series diverges, is there a more straightforward method I am missing? One that does not use $e$ necessarily.




See original post: Series diverges






real-analysis sequences-and-series alternative-proof




real-analysis sequences-and-series alternative-proof






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 28 at 12:11









Wesley Strik

1,514422




1,514422








  • 10




    I would say that proving that the general term does not converge to zero is about as simple as it gets.
    – MisterRiemann
    Nov 28 at 12:17














  • 10




    I would say that proving that the general term does not converge to zero is about as simple as it gets.
    – MisterRiemann
    Nov 28 at 12:17








10




10




I would say that proving that the general term does not converge to zero is about as simple as it gets.
– MisterRiemann
Nov 28 at 12:17




I would say that proving that the general term does not converge to zero is about as simple as it gets.
– MisterRiemann
Nov 28 at 12:17










2 Answers
2






active

oldest

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1














if $n>2$ then $Big(dfrac{n-2}{n}Big)^n>0$ and
$$
\lim_{nto+infty}{Big(frac{n-2}{n}Big)^n}=lim_{nto+infty}{Big(1-frac{2}{n}Big)^n}=e^{-2}>0
$$






share|cite|improve this answer





















  • This is the approach I used in that question, so it does not really answer my question.
    – Wesley Strik
    Nov 28 at 20:06










  • Then why did you accept it?
    – marty cohen
    Nov 28 at 21:18



















0














Suppose we don't know anything about $e$. We can still show that $(1-2/k)^knotto0$ as $ktoinfty$, using the fact that the (alternating) binomial expansion can be truncated after a few terms to give an inequality, since $2^k/k!lt1$ for $kge4$.



$$begin{align}
left(k-2over k right)^k
&=left(1-{2over k}right)^k\
&=1-{kchoose1}left(2over kright)+{kchoose2}left(2over kright)^2-{kchoose3}left(2over kright)^3+cdots\
&=1-2+2-{8over6}{(k-1)(k-2)over k^2}+{16over24}{(k-1)(k-2)(k-3)over k^3}-{32over120}{(k-1)(k-2)(k-3)(k-4)over k^4}+cdots\
&gt1-2+2-{4over3}+{2over3}left(1-{6over k}right)-{4over15}\
&={1over15}-{4over k}
end{align}$$



A nice way to wrap things up is now to note that, for $kge105$, we have



$$left(k-2over kright)^kgt{1over15}-{4over105}={1over35}$$



and therefore $(1-2/k)^knotto0$ as $ktoinfty$, which in turn implies $sum(1-2/k)^k$ diverges.



Remarks: In addition to $2^k/k!lt1$ for $kge4$, which permitted truncation after a subtraction, we used the crude inequalities $(k-1)(k-2)lt k^2$ and $(k-1)(k-2)(k-3)(k-4)lt k^4$ and the slightly less crude inequality $(k-1)(k-2)(k-3)=k^3-6k^2+11k-6gt k^3-6k^2$.






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    2 Answers
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    1














    if $n>2$ then $Big(dfrac{n-2}{n}Big)^n>0$ and
    $$
    \lim_{nto+infty}{Big(frac{n-2}{n}Big)^n}=lim_{nto+infty}{Big(1-frac{2}{n}Big)^n}=e^{-2}>0
    $$






    share|cite|improve this answer





















    • This is the approach I used in that question, so it does not really answer my question.
      – Wesley Strik
      Nov 28 at 20:06










    • Then why did you accept it?
      – marty cohen
      Nov 28 at 21:18
















    1














    if $n>2$ then $Big(dfrac{n-2}{n}Big)^n>0$ and
    $$
    \lim_{nto+infty}{Big(frac{n-2}{n}Big)^n}=lim_{nto+infty}{Big(1-frac{2}{n}Big)^n}=e^{-2}>0
    $$






    share|cite|improve this answer





















    • This is the approach I used in that question, so it does not really answer my question.
      – Wesley Strik
      Nov 28 at 20:06










    • Then why did you accept it?
      – marty cohen
      Nov 28 at 21:18














    1












    1








    1






    if $n>2$ then $Big(dfrac{n-2}{n}Big)^n>0$ and
    $$
    \lim_{nto+infty}{Big(frac{n-2}{n}Big)^n}=lim_{nto+infty}{Big(1-frac{2}{n}Big)^n}=e^{-2}>0
    $$






    share|cite|improve this answer












    if $n>2$ then $Big(dfrac{n-2}{n}Big)^n>0$ and
    $$
    \lim_{nto+infty}{Big(frac{n-2}{n}Big)^n}=lim_{nto+infty}{Big(1-frac{2}{n}Big)^n}=e^{-2}>0
    $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 28 at 20:01









    Samvel Safaryan

    511111




    511111












    • This is the approach I used in that question, so it does not really answer my question.
      – Wesley Strik
      Nov 28 at 20:06










    • Then why did you accept it?
      – marty cohen
      Nov 28 at 21:18


















    • This is the approach I used in that question, so it does not really answer my question.
      – Wesley Strik
      Nov 28 at 20:06










    • Then why did you accept it?
      – marty cohen
      Nov 28 at 21:18
















    This is the approach I used in that question, so it does not really answer my question.
    – Wesley Strik
    Nov 28 at 20:06




    This is the approach I used in that question, so it does not really answer my question.
    – Wesley Strik
    Nov 28 at 20:06












    Then why did you accept it?
    – marty cohen
    Nov 28 at 21:18




    Then why did you accept it?
    – marty cohen
    Nov 28 at 21:18











    0














    Suppose we don't know anything about $e$. We can still show that $(1-2/k)^knotto0$ as $ktoinfty$, using the fact that the (alternating) binomial expansion can be truncated after a few terms to give an inequality, since $2^k/k!lt1$ for $kge4$.



    $$begin{align}
    left(k-2over k right)^k
    &=left(1-{2over k}right)^k\
    &=1-{kchoose1}left(2over kright)+{kchoose2}left(2over kright)^2-{kchoose3}left(2over kright)^3+cdots\
    &=1-2+2-{8over6}{(k-1)(k-2)over k^2}+{16over24}{(k-1)(k-2)(k-3)over k^3}-{32over120}{(k-1)(k-2)(k-3)(k-4)over k^4}+cdots\
    &gt1-2+2-{4over3}+{2over3}left(1-{6over k}right)-{4over15}\
    &={1over15}-{4over k}
    end{align}$$



    A nice way to wrap things up is now to note that, for $kge105$, we have



    $$left(k-2over kright)^kgt{1over15}-{4over105}={1over35}$$



    and therefore $(1-2/k)^knotto0$ as $ktoinfty$, which in turn implies $sum(1-2/k)^k$ diverges.



    Remarks: In addition to $2^k/k!lt1$ for $kge4$, which permitted truncation after a subtraction, we used the crude inequalities $(k-1)(k-2)lt k^2$ and $(k-1)(k-2)(k-3)(k-4)lt k^4$ and the slightly less crude inequality $(k-1)(k-2)(k-3)=k^3-6k^2+11k-6gt k^3-6k^2$.






    share|cite|improve this answer


























      0














      Suppose we don't know anything about $e$. We can still show that $(1-2/k)^knotto0$ as $ktoinfty$, using the fact that the (alternating) binomial expansion can be truncated after a few terms to give an inequality, since $2^k/k!lt1$ for $kge4$.



      $$begin{align}
      left(k-2over k right)^k
      &=left(1-{2over k}right)^k\
      &=1-{kchoose1}left(2over kright)+{kchoose2}left(2over kright)^2-{kchoose3}left(2over kright)^3+cdots\
      &=1-2+2-{8over6}{(k-1)(k-2)over k^2}+{16over24}{(k-1)(k-2)(k-3)over k^3}-{32over120}{(k-1)(k-2)(k-3)(k-4)over k^4}+cdots\
      &gt1-2+2-{4over3}+{2over3}left(1-{6over k}right)-{4over15}\
      &={1over15}-{4over k}
      end{align}$$



      A nice way to wrap things up is now to note that, for $kge105$, we have



      $$left(k-2over kright)^kgt{1over15}-{4over105}={1over35}$$



      and therefore $(1-2/k)^knotto0$ as $ktoinfty$, which in turn implies $sum(1-2/k)^k$ diverges.



      Remarks: In addition to $2^k/k!lt1$ for $kge4$, which permitted truncation after a subtraction, we used the crude inequalities $(k-1)(k-2)lt k^2$ and $(k-1)(k-2)(k-3)(k-4)lt k^4$ and the slightly less crude inequality $(k-1)(k-2)(k-3)=k^3-6k^2+11k-6gt k^3-6k^2$.






      share|cite|improve this answer
























        0












        0








        0






        Suppose we don't know anything about $e$. We can still show that $(1-2/k)^knotto0$ as $ktoinfty$, using the fact that the (alternating) binomial expansion can be truncated after a few terms to give an inequality, since $2^k/k!lt1$ for $kge4$.



        $$begin{align}
        left(k-2over k right)^k
        &=left(1-{2over k}right)^k\
        &=1-{kchoose1}left(2over kright)+{kchoose2}left(2over kright)^2-{kchoose3}left(2over kright)^3+cdots\
        &=1-2+2-{8over6}{(k-1)(k-2)over k^2}+{16over24}{(k-1)(k-2)(k-3)over k^3}-{32over120}{(k-1)(k-2)(k-3)(k-4)over k^4}+cdots\
        &gt1-2+2-{4over3}+{2over3}left(1-{6over k}right)-{4over15}\
        &={1over15}-{4over k}
        end{align}$$



        A nice way to wrap things up is now to note that, for $kge105$, we have



        $$left(k-2over kright)^kgt{1over15}-{4over105}={1over35}$$



        and therefore $(1-2/k)^knotto0$ as $ktoinfty$, which in turn implies $sum(1-2/k)^k$ diverges.



        Remarks: In addition to $2^k/k!lt1$ for $kge4$, which permitted truncation after a subtraction, we used the crude inequalities $(k-1)(k-2)lt k^2$ and $(k-1)(k-2)(k-3)(k-4)lt k^4$ and the slightly less crude inequality $(k-1)(k-2)(k-3)=k^3-6k^2+11k-6gt k^3-6k^2$.






        share|cite|improve this answer












        Suppose we don't know anything about $e$. We can still show that $(1-2/k)^knotto0$ as $ktoinfty$, using the fact that the (alternating) binomial expansion can be truncated after a few terms to give an inequality, since $2^k/k!lt1$ for $kge4$.



        $$begin{align}
        left(k-2over k right)^k
        &=left(1-{2over k}right)^k\
        &=1-{kchoose1}left(2over kright)+{kchoose2}left(2over kright)^2-{kchoose3}left(2over kright)^3+cdots\
        &=1-2+2-{8over6}{(k-1)(k-2)over k^2}+{16over24}{(k-1)(k-2)(k-3)over k^3}-{32over120}{(k-1)(k-2)(k-3)(k-4)over k^4}+cdots\
        &gt1-2+2-{4over3}+{2over3}left(1-{6over k}right)-{4over15}\
        &={1over15}-{4over k}
        end{align}$$



        A nice way to wrap things up is now to note that, for $kge105$, we have



        $$left(k-2over kright)^kgt{1over15}-{4over105}={1over35}$$



        and therefore $(1-2/k)^knotto0$ as $ktoinfty$, which in turn implies $sum(1-2/k)^k$ diverges.



        Remarks: In addition to $2^k/k!lt1$ for $kge4$, which permitted truncation after a subtraction, we used the crude inequalities $(k-1)(k-2)lt k^2$ and $(k-1)(k-2)(k-3)(k-4)lt k^4$ and the slightly less crude inequality $(k-1)(k-2)(k-3)=k^3-6k^2+11k-6gt k^3-6k^2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 at 22:46









        Barry Cipra

        58.8k653123




        58.8k653123






























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