Converting summation ( how do I get from $k cdot sum_{j=k}^{2k} frac{1}{j-k+1}$ to $ k cdot sum_{j=1}^{k+1}...












1












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I am trying to figure out the steps between these two equal expressions.



$$ k cdot sum_{j=k}^{2k} frac{1}{j-k+1} = k cdot sum_{j=1}^{k+1} frac{1}{j} $$



I get that $2k -k +1 = k+1$, but why does $j=k$ change to $j=1$?






discrete-mathematics summation







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  • 1




    $begingroup$
    Because $k-k+1=1$ by using the same exact argument as for $2k$
    $endgroup$
    – b00n heT
    Dec 17 '18 at 21:33








  • 1




    $begingroup$
    And the $k$ in front of the sum is irrelevant. You might as well divide that out...
    $endgroup$
    – Mason
    Dec 17 '18 at 21:35
















1












$begingroup$


I am trying to figure out the steps between these two equal expressions.



$$ k cdot sum_{j=k}^{2k} frac{1}{j-k+1} = k cdot sum_{j=1}^{k+1} frac{1}{j} $$



I get that $2k -k +1 = k+1$, but why does $j=k$ change to $j=1$?






discrete-mathematics summation







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Because $k-k+1=1$ by using the same exact argument as for $2k$
    $endgroup$
    – b00n heT
    Dec 17 '18 at 21:33








  • 1




    $begingroup$
    And the $k$ in front of the sum is irrelevant. You might as well divide that out...
    $endgroup$
    – Mason
    Dec 17 '18 at 21:35














1












1








1





$begingroup$


I am trying to figure out the steps between these two equal expressions.



$$ k cdot sum_{j=k}^{2k} frac{1}{j-k+1} = k cdot sum_{j=1}^{k+1} frac{1}{j} $$



I get that $2k -k +1 = k+1$, but why does $j=k$ change to $j=1$?






discrete-mathematics summation







share|cite|improve this question











$endgroup$




I am trying to figure out the steps between these two equal expressions.



$$ k cdot sum_{j=k}^{2k} frac{1}{j-k+1} = k cdot sum_{j=1}^{k+1} frac{1}{j} $$



I get that $2k -k +1 = k+1$, but why does $j=k$ change to $j=1$?






discrete-mathematics summation




discrete-mathematics summation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 21:36









Martin Argerami

128k1184184




128k1184184










asked Dec 17 '18 at 21:26









LuckyLucky

1177




1177








  • 1




    $begingroup$
    Because $k-k+1=1$ by using the same exact argument as for $2k$
    $endgroup$
    – b00n heT
    Dec 17 '18 at 21:33








  • 1




    $begingroup$
    And the $k$ in front of the sum is irrelevant. You might as well divide that out...
    $endgroup$
    – Mason
    Dec 17 '18 at 21:35














  • 1




    $begingroup$
    Because $k-k+1=1$ by using the same exact argument as for $2k$
    $endgroup$
    – b00n heT
    Dec 17 '18 at 21:33








  • 1




    $begingroup$
    And the $k$ in front of the sum is irrelevant. You might as well divide that out...
    $endgroup$
    – Mason
    Dec 17 '18 at 21:35








1




1




$begingroup$
Because $k-k+1=1$ by using the same exact argument as for $2k$
$endgroup$
– b00n heT
Dec 17 '18 at 21:33






$begingroup$
Because $k-k+1=1$ by using the same exact argument as for $2k$
$endgroup$
– b00n heT
Dec 17 '18 at 21:33






1




1




$begingroup$
And the $k$ in front of the sum is irrelevant. You might as well divide that out...
$endgroup$
– Mason
Dec 17 '18 at 21:35




$begingroup$
And the $k$ in front of the sum is irrelevant. You might as well divide that out...
$endgroup$
– Mason
Dec 17 '18 at 21:35










1 Answer
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$begingroup$

Note that for





  • $j$ varing from $k$ up to $2k$


we have that





  • $j-k+1$ varies linearly from $1$ to $k+1$


therefore the two sums are equivalent.



If you find confusing using the same $j$ for both sums, let use another index for the second sum, that is for example



$$r=j-k+1 implies k cdot sum_{j=k}^{2k} frac{1}{j-k+1} = k cdot sum_{r=1}^{k+1} frac{1}{r}$$






share|cite|improve this answer











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    1 Answer
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    active

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    1 Answer
    1






    active

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    active

    oldest

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    active

    oldest

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    2












    $begingroup$

    Note that for





    • $j$ varing from $k$ up to $2k$


    we have that





    • $j-k+1$ varies linearly from $1$ to $k+1$


    therefore the two sums are equivalent.



    If you find confusing using the same $j$ for both sums, let use another index for the second sum, that is for example



    $$r=j-k+1 implies k cdot sum_{j=k}^{2k} frac{1}{j-k+1} = k cdot sum_{r=1}^{k+1} frac{1}{r}$$






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Note that for





      • $j$ varing from $k$ up to $2k$


      we have that





      • $j-k+1$ varies linearly from $1$ to $k+1$


      therefore the two sums are equivalent.



      If you find confusing using the same $j$ for both sums, let use another index for the second sum, that is for example



      $$r=j-k+1 implies k cdot sum_{j=k}^{2k} frac{1}{j-k+1} = k cdot sum_{r=1}^{k+1} frac{1}{r}$$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Note that for





        • $j$ varing from $k$ up to $2k$


        we have that





        • $j-k+1$ varies linearly from $1$ to $k+1$


        therefore the two sums are equivalent.



        If you find confusing using the same $j$ for both sums, let use another index for the second sum, that is for example



        $$r=j-k+1 implies k cdot sum_{j=k}^{2k} frac{1}{j-k+1} = k cdot sum_{r=1}^{k+1} frac{1}{r}$$






        share|cite|improve this answer











        $endgroup$



        Note that for





        • $j$ varing from $k$ up to $2k$


        we have that





        • $j-k+1$ varies linearly from $1$ to $k+1$


        therefore the two sums are equivalent.



        If you find confusing using the same $j$ for both sums, let use another index for the second sum, that is for example



        $$r=j-k+1 implies k cdot sum_{j=k}^{2k} frac{1}{j-k+1} = k cdot sum_{r=1}^{k+1} frac{1}{r}$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 17 '18 at 21:34

























        answered Dec 17 '18 at 21:29









        gimusigimusi

        93k84594




        93k84594






























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