Evaluate $sumlimits_{r=1}^{50}left[frac{1}{49+r} - frac{1}{2r(2r-1)}right]$












0












$begingroup$


Find
$$
S=sum_{r=1}^{50}left[frac{1}{49+r} - frac{1}{2r(2r-1)}right]
$$
Thus, some terms are in harmonic progression.



I tried to rearrange $S$ and represent it as a sum of two terms.






sequences-and-series summation







share|cite|improve this question











$endgroup$












  • $begingroup$
    So what is $HP$, and why wouldn't you spell it out anyway?
    $endgroup$
    – David G. Stork
    May 3 '18 at 17:41








  • 1




    $begingroup$
    @DavidG.Stork I believe it's harmonic progression.
    $endgroup$
    – Karn Watcharasupat
    May 3 '18 at 18:07






  • 1




    $begingroup$
    The answer is $$frac{1}{100}$$ by the way. Still figuring out the working...
    $endgroup$
    – Karn Watcharasupat
    May 3 '18 at 18:21










  • $begingroup$
    You need a $sum$ symbol on the second one$ldots$
    $endgroup$
    – Felix Marin
    May 3 '18 at 18:39








  • 2




    $begingroup$
    $New$ user info: Please, you must always show what you tried. That enhances the possibility that MSE-people pay attention to your post. In addition, read the MSE-$LaTeX$-$texttt{MathJax}$ Tutorial.
    $endgroup$
    – Felix Marin
    May 3 '18 at 18:45


















0












$begingroup$


Find
$$
S=sum_{r=1}^{50}left[frac{1}{49+r} - frac{1}{2r(2r-1)}right]
$$
Thus, some terms are in harmonic progression.



I tried to rearrange $S$ and represent it as a sum of two terms.






sequences-and-series summation







share|cite|improve this question











$endgroup$












  • $begingroup$
    So what is $HP$, and why wouldn't you spell it out anyway?
    $endgroup$
    – David G. Stork
    May 3 '18 at 17:41








  • 1




    $begingroup$
    @DavidG.Stork I believe it's harmonic progression.
    $endgroup$
    – Karn Watcharasupat
    May 3 '18 at 18:07






  • 1




    $begingroup$
    The answer is $$frac{1}{100}$$ by the way. Still figuring out the working...
    $endgroup$
    – Karn Watcharasupat
    May 3 '18 at 18:21










  • $begingroup$
    You need a $sum$ symbol on the second one$ldots$
    $endgroup$
    – Felix Marin
    May 3 '18 at 18:39








  • 2




    $begingroup$
    $New$ user info: Please, you must always show what you tried. That enhances the possibility that MSE-people pay attention to your post. In addition, read the MSE-$LaTeX$-$texttt{MathJax}$ Tutorial.
    $endgroup$
    – Felix Marin
    May 3 '18 at 18:45
















0












0








0


2



$begingroup$


Find
$$
S=sum_{r=1}^{50}left[frac{1}{49+r} - frac{1}{2r(2r-1)}right]
$$
Thus, some terms are in harmonic progression.



I tried to rearrange $S$ and represent it as a sum of two terms.






sequences-and-series summation







share|cite|improve this question











$endgroup$




Find
$$
S=sum_{r=1}^{50}left[frac{1}{49+r} - frac{1}{2r(2r-1)}right]
$$
Thus, some terms are in harmonic progression.



I tried to rearrange $S$ and represent it as a sum of two terms.






sequences-and-series summation




sequences-and-series summation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 11:49









Did

248k23223460




248k23223460










asked May 3 '18 at 17:40









Dikshit GautamDikshit Gautam

795




795












  • $begingroup$
    So what is $HP$, and why wouldn't you spell it out anyway?
    $endgroup$
    – David G. Stork
    May 3 '18 at 17:41








  • 1




    $begingroup$
    @DavidG.Stork I believe it's harmonic progression.
    $endgroup$
    – Karn Watcharasupat
    May 3 '18 at 18:07






  • 1




    $begingroup$
    The answer is $$frac{1}{100}$$ by the way. Still figuring out the working...
    $endgroup$
    – Karn Watcharasupat
    May 3 '18 at 18:21










  • $begingroup$
    You need a $sum$ symbol on the second one$ldots$
    $endgroup$
    – Felix Marin
    May 3 '18 at 18:39








  • 2




    $begingroup$
    $New$ user info: Please, you must always show what you tried. That enhances the possibility that MSE-people pay attention to your post. In addition, read the MSE-$LaTeX$-$texttt{MathJax}$ Tutorial.
    $endgroup$
    – Felix Marin
    May 3 '18 at 18:45




















  • $begingroup$
    So what is $HP$, and why wouldn't you spell it out anyway?
    $endgroup$
    – David G. Stork
    May 3 '18 at 17:41








  • 1




    $begingroup$
    @DavidG.Stork I believe it's harmonic progression.
    $endgroup$
    – Karn Watcharasupat
    May 3 '18 at 18:07






  • 1




    $begingroup$
    The answer is $$frac{1}{100}$$ by the way. Still figuring out the working...
    $endgroup$
    – Karn Watcharasupat
    May 3 '18 at 18:21










  • $begingroup$
    You need a $sum$ symbol on the second one$ldots$
    $endgroup$
    – Felix Marin
    May 3 '18 at 18:39








  • 2




    $begingroup$
    $New$ user info: Please, you must always show what you tried. That enhances the possibility that MSE-people pay attention to your post. In addition, read the MSE-$LaTeX$-$texttt{MathJax}$ Tutorial.
    $endgroup$
    – Felix Marin
    May 3 '18 at 18:45


















$begingroup$
So what is $HP$, and why wouldn't you spell it out anyway?
$endgroup$
– David G. Stork
May 3 '18 at 17:41






$begingroup$
So what is $HP$, and why wouldn't you spell it out anyway?
$endgroup$
– David G. Stork
May 3 '18 at 17:41






1




1




$begingroup$
@DavidG.Stork I believe it's harmonic progression.
$endgroup$
– Karn Watcharasupat
May 3 '18 at 18:07




$begingroup$
@DavidG.Stork I believe it's harmonic progression.
$endgroup$
– Karn Watcharasupat
May 3 '18 at 18:07




1




1




$begingroup$
The answer is $$frac{1}{100}$$ by the way. Still figuring out the working...
$endgroup$
– Karn Watcharasupat
May 3 '18 at 18:21




$begingroup$
The answer is $$frac{1}{100}$$ by the way. Still figuring out the working...
$endgroup$
– Karn Watcharasupat
May 3 '18 at 18:21












$begingroup$
You need a $sum$ symbol on the second one$ldots$
$endgroup$
– Felix Marin
May 3 '18 at 18:39






$begingroup$
You need a $sum$ symbol on the second one$ldots$
$endgroup$
– Felix Marin
May 3 '18 at 18:39






2




2




$begingroup$
$New$ user info: Please, you must always show what you tried. That enhances the possibility that MSE-people pay attention to your post. In addition, read the MSE-$LaTeX$-$texttt{MathJax}$ Tutorial.
$endgroup$
– Felix Marin
May 3 '18 at 18:45






$begingroup$
$New$ user info: Please, you must always show what you tried. That enhances the possibility that MSE-people pay attention to your post. In addition, read the MSE-$LaTeX$-$texttt{MathJax}$ Tutorial.
$endgroup$
– Felix Marin
May 3 '18 at 18:45












1 Answer
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$begingroup$

$S=sumlimits_{r=1}^{50}left[frac{1}{49+r} - frac{1}{2r(2r-1)}right]$



The first part of the sum:



$sumlimits_{r=1}^{50}frac{1}{49+r}=sumlimits_{r=50}^{99}frac{1}{r}=sumlimits_{r=1}^{99}frac{1}{r}-sumlimits_{r=1}^{49}frac{1}{r}$



The second part of the sum:



$sumlimits_{r=1}^{50} - frac{1}{2r(2r-1)}=sumlimits_{r=1}^{50}big( frac{1}{2r}{-frac{1}{(2r-1)}}big)=frac{1}{100}-sumlimits_{r=1}^{99}frac{(-1)^{r-1}}{r}$



Finally



$S=frac{1}{100}+sumlimits_{r=1}^{99}big(frac{1}{r}-frac{(-1)^{r-1}}{r}big)-sumlimits_{r=1}^{49}frac{1}{r}$ as the 99. term is equal to zero in the first part of the sum and $r=2k$ we get the followings:



$S=frac{1}{100}+sumlimits_{k=1}^{49}frac{2}{2k}-sumlimits_{r=1}^{49}frac{1}{r}=frac{1}{100}$






share|cite|improve this answer









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    1












    $begingroup$

    $S=sumlimits_{r=1}^{50}left[frac{1}{49+r} - frac{1}{2r(2r-1)}right]$



    The first part of the sum:



    $sumlimits_{r=1}^{50}frac{1}{49+r}=sumlimits_{r=50}^{99}frac{1}{r}=sumlimits_{r=1}^{99}frac{1}{r}-sumlimits_{r=1}^{49}frac{1}{r}$



    The second part of the sum:



    $sumlimits_{r=1}^{50} - frac{1}{2r(2r-1)}=sumlimits_{r=1}^{50}big( frac{1}{2r}{-frac{1}{(2r-1)}}big)=frac{1}{100}-sumlimits_{r=1}^{99}frac{(-1)^{r-1}}{r}$



    Finally



    $S=frac{1}{100}+sumlimits_{r=1}^{99}big(frac{1}{r}-frac{(-1)^{r-1}}{r}big)-sumlimits_{r=1}^{49}frac{1}{r}$ as the 99. term is equal to zero in the first part of the sum and $r=2k$ we get the followings:



    $S=frac{1}{100}+sumlimits_{k=1}^{49}frac{2}{2k}-sumlimits_{r=1}^{49}frac{1}{r}=frac{1}{100}$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $S=sumlimits_{r=1}^{50}left[frac{1}{49+r} - frac{1}{2r(2r-1)}right]$



      The first part of the sum:



      $sumlimits_{r=1}^{50}frac{1}{49+r}=sumlimits_{r=50}^{99}frac{1}{r}=sumlimits_{r=1}^{99}frac{1}{r}-sumlimits_{r=1}^{49}frac{1}{r}$



      The second part of the sum:



      $sumlimits_{r=1}^{50} - frac{1}{2r(2r-1)}=sumlimits_{r=1}^{50}big( frac{1}{2r}{-frac{1}{(2r-1)}}big)=frac{1}{100}-sumlimits_{r=1}^{99}frac{(-1)^{r-1}}{r}$



      Finally



      $S=frac{1}{100}+sumlimits_{r=1}^{99}big(frac{1}{r}-frac{(-1)^{r-1}}{r}big)-sumlimits_{r=1}^{49}frac{1}{r}$ as the 99. term is equal to zero in the first part of the sum and $r=2k$ we get the followings:



      $S=frac{1}{100}+sumlimits_{k=1}^{49}frac{2}{2k}-sumlimits_{r=1}^{49}frac{1}{r}=frac{1}{100}$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $S=sumlimits_{r=1}^{50}left[frac{1}{49+r} - frac{1}{2r(2r-1)}right]$



        The first part of the sum:



        $sumlimits_{r=1}^{50}frac{1}{49+r}=sumlimits_{r=50}^{99}frac{1}{r}=sumlimits_{r=1}^{99}frac{1}{r}-sumlimits_{r=1}^{49}frac{1}{r}$



        The second part of the sum:



        $sumlimits_{r=1}^{50} - frac{1}{2r(2r-1)}=sumlimits_{r=1}^{50}big( frac{1}{2r}{-frac{1}{(2r-1)}}big)=frac{1}{100}-sumlimits_{r=1}^{99}frac{(-1)^{r-1}}{r}$



        Finally



        $S=frac{1}{100}+sumlimits_{r=1}^{99}big(frac{1}{r}-frac{(-1)^{r-1}}{r}big)-sumlimits_{r=1}^{49}frac{1}{r}$ as the 99. term is equal to zero in the first part of the sum and $r=2k$ we get the followings:



        $S=frac{1}{100}+sumlimits_{k=1}^{49}frac{2}{2k}-sumlimits_{r=1}^{49}frac{1}{r}=frac{1}{100}$






        share|cite|improve this answer









        $endgroup$



        $S=sumlimits_{r=1}^{50}left[frac{1}{49+r} - frac{1}{2r(2r-1)}right]$



        The first part of the sum:



        $sumlimits_{r=1}^{50}frac{1}{49+r}=sumlimits_{r=50}^{99}frac{1}{r}=sumlimits_{r=1}^{99}frac{1}{r}-sumlimits_{r=1}^{49}frac{1}{r}$



        The second part of the sum:



        $sumlimits_{r=1}^{50} - frac{1}{2r(2r-1)}=sumlimits_{r=1}^{50}big( frac{1}{2r}{-frac{1}{(2r-1)}}big)=frac{1}{100}-sumlimits_{r=1}^{99}frac{(-1)^{r-1}}{r}$



        Finally



        $S=frac{1}{100}+sumlimits_{r=1}^{99}big(frac{1}{r}-frac{(-1)^{r-1}}{r}big)-sumlimits_{r=1}^{49}frac{1}{r}$ as the 99. term is equal to zero in the first part of the sum and $r=2k$ we get the followings:



        $S=frac{1}{100}+sumlimits_{k=1}^{49}frac{2}{2k}-sumlimits_{r=1}^{49}frac{1}{r}=frac{1}{100}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered May 4 '18 at 13:39









        JV.StalkerJV.Stalker

        68339




        68339






























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