A Question based on series.












1














Find the sum of$$sum _{r=1}^inftyleft({frac{2}{4r-3}-frac{1}{2r}}right)$$ I tried to solve this problem by taking $2$ common from the expression and got the result as of$$2sum _{r=1}^inftyleft({frac{1}{4r-3}-frac{1}{4r}}right)$$ but I could not get any constant term on which I could apply the telescopic series formula please help me out.






sequences-and-series telescopic-series







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  • Not sure if this helps, but seems somewhat similar to series for $frac pi 4$: en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80#Convergence
    – Noble Mushtak
    Dec 29 '18 at 1:22
















1














Find the sum of$$sum _{r=1}^inftyleft({frac{2}{4r-3}-frac{1}{2r}}right)$$ I tried to solve this problem by taking $2$ common from the expression and got the result as of$$2sum _{r=1}^inftyleft({frac{1}{4r-3}-frac{1}{4r}}right)$$ but I could not get any constant term on which I could apply the telescopic series formula please help me out.






sequences-and-series telescopic-series







share|cite|improve this question
























  • Not sure if this helps, but seems somewhat similar to series for $frac pi 4$: en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80#Convergence
    – Noble Mushtak
    Dec 29 '18 at 1:22














1












1








1


1





Find the sum of$$sum _{r=1}^inftyleft({frac{2}{4r-3}-frac{1}{2r}}right)$$ I tried to solve this problem by taking $2$ common from the expression and got the result as of$$2sum _{r=1}^inftyleft({frac{1}{4r-3}-frac{1}{4r}}right)$$ but I could not get any constant term on which I could apply the telescopic series formula please help me out.






sequences-and-series telescopic-series







share|cite|improve this question















Find the sum of$$sum _{r=1}^inftyleft({frac{2}{4r-3}-frac{1}{2r}}right)$$ I tried to solve this problem by taking $2$ common from the expression and got the result as of$$2sum _{r=1}^inftyleft({frac{1}{4r-3}-frac{1}{4r}}right)$$ but I could not get any constant term on which I could apply the telescopic series formula please help me out.






sequences-and-series telescopic-series




sequences-and-series telescopic-series






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edited Dec 29 '18 at 9:51









Ivan Neretin

8,83821535




8,83821535










asked Dec 29 '18 at 1:09









Bipul KumarBipul Kumar

355




355












  • Not sure if this helps, but seems somewhat similar to series for $frac pi 4$: en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80#Convergence
    – Noble Mushtak
    Dec 29 '18 at 1:22


















  • Not sure if this helps, but seems somewhat similar to series for $frac pi 4$: en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80#Convergence
    – Noble Mushtak
    Dec 29 '18 at 1:22
















Not sure if this helps, but seems somewhat similar to series for $frac pi 4$: en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80#Convergence
– Noble Mushtak
Dec 29 '18 at 1:22




Not sure if this helps, but seems somewhat similar to series for $frac pi 4$: en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80#Convergence
– Noble Mushtak
Dec 29 '18 at 1:22










2 Answers
2






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4














This is not a telescopic series, though it might superficially resemble one.



The common technique with such problems is to transform them to power series (by putting $x^n$ or something in the numerator), then fiddle around with functions, their Taylor series, maybe some differentiation or other stuff, and then evaluate the series at $x=1$.



Short of that, we are reduced to using the following crude tools.



Suppose you know that $sumlimits_{n=1}^infty{(-1)^{n+1}over n}=ln2$ (because if you don't, we aren't going anywhere). Suppose you also know that $sumlimits_{n=1}^infty{(-1)^{n+1}over 2n+1}={piover4}$, which is also a mandatory prerequisite. Then, obviously
$$sum_{n=1}^infty{(-1)^{n+1}over 2n}={1over2}sum_{n=1}^infty{(-1)^{n+1}over n}={ln2over2}$$
Now I'm going to put it in a simple way, without all those fancy $sum$s:
$$begin{array}{ccccccccccc}
{piover4} &=& 1 & & -{1over3} & & +{1over5} & & -{1over7} &dots \
ln 2 &=& 1 & -{1over2} & +{1over3} & -{1over4} & +{1over5} & -{1over6} & +{1over7} &dots \
{ln 2over2} &=& & phantom{-}{1over2} & & -{1over4} & & +{1over6} & &dots \
end{array}$$
Now just add these three series together.






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    -1














    $$S_n=sum _{r=1}^n left(frac{2}{4r-3}-frac{1}{2r}right)=frac12sum _{r=1}^n left(frac{1}{r+frac14-1}-frac{1}{r+1-1}right)$$
    It is a easy problem to people familiar with the polygamma functions.
    $$sum _{r=1}^n frac{1}{r+a-1}=psi(a+n)-psi(a)$$
    $psi(x)$ is the digamma function.



    Successively with $a=frac14$ and $a=1$ :
    $$S_n=frac12left(psileft(frac14+nright)-psileft(frac14right)-psileft(1+nright)+psileft(1right)right)$$
    $ntoinftyquad psi(a+n)sim ln(n)+Oleft(frac{1}{x}right)$



    $lim_{ntoinfty}left(psileft(frac14+nright)-psileft(1+nright)right)=0$
    $$S_infty=frac12left(-psileft(-frac14right)+psileft(1right)right)$$
    $psi(1)=-gamma$ Euler-Mascheroni constant. And $psi(1/4)=-gamma-frac{pi}{2}-3ln(2)$
    $$S_infty=frac12left(frac{pi}{2}+3ln(2)right)$$
    $$sum _{r=1}^infty left(frac{2}{4r-3}-frac{1}{2r}right)=frac{pi}{4}+frac32ln(2)$$
    For reference : http://mathworld.wolfram.com/DigammaFunction.html






    share|cite|improve this answer





















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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      4














      This is not a telescopic series, though it might superficially resemble one.



      The common technique with such problems is to transform them to power series (by putting $x^n$ or something in the numerator), then fiddle around with functions, their Taylor series, maybe some differentiation or other stuff, and then evaluate the series at $x=1$.



      Short of that, we are reduced to using the following crude tools.



      Suppose you know that $sumlimits_{n=1}^infty{(-1)^{n+1}over n}=ln2$ (because if you don't, we aren't going anywhere). Suppose you also know that $sumlimits_{n=1}^infty{(-1)^{n+1}over 2n+1}={piover4}$, which is also a mandatory prerequisite. Then, obviously
      $$sum_{n=1}^infty{(-1)^{n+1}over 2n}={1over2}sum_{n=1}^infty{(-1)^{n+1}over n}={ln2over2}$$
      Now I'm going to put it in a simple way, without all those fancy $sum$s:
      $$begin{array}{ccccccccccc}
      {piover4} &=& 1 & & -{1over3} & & +{1over5} & & -{1over7} &dots \
      ln 2 &=& 1 & -{1over2} & +{1over3} & -{1over4} & +{1over5} & -{1over6} & +{1over7} &dots \
      {ln 2over2} &=& & phantom{-}{1over2} & & -{1over4} & & +{1over6} & &dots \
      end{array}$$
      Now just add these three series together.






      share|cite|improve this answer




























        4














        This is not a telescopic series, though it might superficially resemble one.



        The common technique with such problems is to transform them to power series (by putting $x^n$ or something in the numerator), then fiddle around with functions, their Taylor series, maybe some differentiation or other stuff, and then evaluate the series at $x=1$.



        Short of that, we are reduced to using the following crude tools.



        Suppose you know that $sumlimits_{n=1}^infty{(-1)^{n+1}over n}=ln2$ (because if you don't, we aren't going anywhere). Suppose you also know that $sumlimits_{n=1}^infty{(-1)^{n+1}over 2n+1}={piover4}$, which is also a mandatory prerequisite. Then, obviously
        $$sum_{n=1}^infty{(-1)^{n+1}over 2n}={1over2}sum_{n=1}^infty{(-1)^{n+1}over n}={ln2over2}$$
        Now I'm going to put it in a simple way, without all those fancy $sum$s:
        $$begin{array}{ccccccccccc}
        {piover4} &=& 1 & & -{1over3} & & +{1over5} & & -{1over7} &dots \
        ln 2 &=& 1 & -{1over2} & +{1over3} & -{1over4} & +{1over5} & -{1over6} & +{1over7} &dots \
        {ln 2over2} &=& & phantom{-}{1over2} & & -{1over4} & & +{1over6} & &dots \
        end{array}$$
        Now just add these three series together.






        share|cite|improve this answer


























          4












          4








          4






          This is not a telescopic series, though it might superficially resemble one.



          The common technique with such problems is to transform them to power series (by putting $x^n$ or something in the numerator), then fiddle around with functions, their Taylor series, maybe some differentiation or other stuff, and then evaluate the series at $x=1$.



          Short of that, we are reduced to using the following crude tools.



          Suppose you know that $sumlimits_{n=1}^infty{(-1)^{n+1}over n}=ln2$ (because if you don't, we aren't going anywhere). Suppose you also know that $sumlimits_{n=1}^infty{(-1)^{n+1}over 2n+1}={piover4}$, which is also a mandatory prerequisite. Then, obviously
          $$sum_{n=1}^infty{(-1)^{n+1}over 2n}={1over2}sum_{n=1}^infty{(-1)^{n+1}over n}={ln2over2}$$
          Now I'm going to put it in a simple way, without all those fancy $sum$s:
          $$begin{array}{ccccccccccc}
          {piover4} &=& 1 & & -{1over3} & & +{1over5} & & -{1over7} &dots \
          ln 2 &=& 1 & -{1over2} & +{1over3} & -{1over4} & +{1over5} & -{1over6} & +{1over7} &dots \
          {ln 2over2} &=& & phantom{-}{1over2} & & -{1over4} & & +{1over6} & &dots \
          end{array}$$
          Now just add these three series together.






          share|cite|improve this answer














          This is not a telescopic series, though it might superficially resemble one.



          The common technique with such problems is to transform them to power series (by putting $x^n$ or something in the numerator), then fiddle around with functions, their Taylor series, maybe some differentiation or other stuff, and then evaluate the series at $x=1$.



          Short of that, we are reduced to using the following crude tools.



          Suppose you know that $sumlimits_{n=1}^infty{(-1)^{n+1}over n}=ln2$ (because if you don't, we aren't going anywhere). Suppose you also know that $sumlimits_{n=1}^infty{(-1)^{n+1}over 2n+1}={piover4}$, which is also a mandatory prerequisite. Then, obviously
          $$sum_{n=1}^infty{(-1)^{n+1}over 2n}={1over2}sum_{n=1}^infty{(-1)^{n+1}over n}={ln2over2}$$
          Now I'm going to put it in a simple way, without all those fancy $sum$s:
          $$begin{array}{ccccccccccc}
          {piover4} &=& 1 & & -{1over3} & & +{1over5} & & -{1over7} &dots \
          ln 2 &=& 1 & -{1over2} & +{1over3} & -{1over4} & +{1over5} & -{1over6} & +{1over7} &dots \
          {ln 2over2} &=& & phantom{-}{1over2} & & -{1over4} & & +{1over6} & &dots \
          end{array}$$
          Now just add these three series together.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 29 '18 at 9:42

























          answered Dec 29 '18 at 9:28









          Ivan NeretinIvan Neretin

          8,83821535




          8,83821535























              -1














              $$S_n=sum _{r=1}^n left(frac{2}{4r-3}-frac{1}{2r}right)=frac12sum _{r=1}^n left(frac{1}{r+frac14-1}-frac{1}{r+1-1}right)$$
              It is a easy problem to people familiar with the polygamma functions.
              $$sum _{r=1}^n frac{1}{r+a-1}=psi(a+n)-psi(a)$$
              $psi(x)$ is the digamma function.



              Successively with $a=frac14$ and $a=1$ :
              $$S_n=frac12left(psileft(frac14+nright)-psileft(frac14right)-psileft(1+nright)+psileft(1right)right)$$
              $ntoinftyquad psi(a+n)sim ln(n)+Oleft(frac{1}{x}right)$



              $lim_{ntoinfty}left(psileft(frac14+nright)-psileft(1+nright)right)=0$
              $$S_infty=frac12left(-psileft(-frac14right)+psileft(1right)right)$$
              $psi(1)=-gamma$ Euler-Mascheroni constant. And $psi(1/4)=-gamma-frac{pi}{2}-3ln(2)$
              $$S_infty=frac12left(frac{pi}{2}+3ln(2)right)$$
              $$sum _{r=1}^infty left(frac{2}{4r-3}-frac{1}{2r}right)=frac{pi}{4}+frac32ln(2)$$
              For reference : http://mathworld.wolfram.com/DigammaFunction.html






              share|cite|improve this answer


























                -1














                $$S_n=sum _{r=1}^n left(frac{2}{4r-3}-frac{1}{2r}right)=frac12sum _{r=1}^n left(frac{1}{r+frac14-1}-frac{1}{r+1-1}right)$$
                It is a easy problem to people familiar with the polygamma functions.
                $$sum _{r=1}^n frac{1}{r+a-1}=psi(a+n)-psi(a)$$
                $psi(x)$ is the digamma function.



                Successively with $a=frac14$ and $a=1$ :
                $$S_n=frac12left(psileft(frac14+nright)-psileft(frac14right)-psileft(1+nright)+psileft(1right)right)$$
                $ntoinftyquad psi(a+n)sim ln(n)+Oleft(frac{1}{x}right)$



                $lim_{ntoinfty}left(psileft(frac14+nright)-psileft(1+nright)right)=0$
                $$S_infty=frac12left(-psileft(-frac14right)+psileft(1right)right)$$
                $psi(1)=-gamma$ Euler-Mascheroni constant. And $psi(1/4)=-gamma-frac{pi}{2}-3ln(2)$
                $$S_infty=frac12left(frac{pi}{2}+3ln(2)right)$$
                $$sum _{r=1}^infty left(frac{2}{4r-3}-frac{1}{2r}right)=frac{pi}{4}+frac32ln(2)$$
                For reference : http://mathworld.wolfram.com/DigammaFunction.html






                share|cite|improve this answer
























                  -1












                  -1








                  -1






                  $$S_n=sum _{r=1}^n left(frac{2}{4r-3}-frac{1}{2r}right)=frac12sum _{r=1}^n left(frac{1}{r+frac14-1}-frac{1}{r+1-1}right)$$
                  It is a easy problem to people familiar with the polygamma functions.
                  $$sum _{r=1}^n frac{1}{r+a-1}=psi(a+n)-psi(a)$$
                  $psi(x)$ is the digamma function.



                  Successively with $a=frac14$ and $a=1$ :
                  $$S_n=frac12left(psileft(frac14+nright)-psileft(frac14right)-psileft(1+nright)+psileft(1right)right)$$
                  $ntoinftyquad psi(a+n)sim ln(n)+Oleft(frac{1}{x}right)$



                  $lim_{ntoinfty}left(psileft(frac14+nright)-psileft(1+nright)right)=0$
                  $$S_infty=frac12left(-psileft(-frac14right)+psileft(1right)right)$$
                  $psi(1)=-gamma$ Euler-Mascheroni constant. And $psi(1/4)=-gamma-frac{pi}{2}-3ln(2)$
                  $$S_infty=frac12left(frac{pi}{2}+3ln(2)right)$$
                  $$sum _{r=1}^infty left(frac{2}{4r-3}-frac{1}{2r}right)=frac{pi}{4}+frac32ln(2)$$
                  For reference : http://mathworld.wolfram.com/DigammaFunction.html






                  share|cite|improve this answer












                  $$S_n=sum _{r=1}^n left(frac{2}{4r-3}-frac{1}{2r}right)=frac12sum _{r=1}^n left(frac{1}{r+frac14-1}-frac{1}{r+1-1}right)$$
                  It is a easy problem to people familiar with the polygamma functions.
                  $$sum _{r=1}^n frac{1}{r+a-1}=psi(a+n)-psi(a)$$
                  $psi(x)$ is the digamma function.



                  Successively with $a=frac14$ and $a=1$ :
                  $$S_n=frac12left(psileft(frac14+nright)-psileft(frac14right)-psileft(1+nright)+psileft(1right)right)$$
                  $ntoinftyquad psi(a+n)sim ln(n)+Oleft(frac{1}{x}right)$



                  $lim_{ntoinfty}left(psileft(frac14+nright)-psileft(1+nright)right)=0$
                  $$S_infty=frac12left(-psileft(-frac14right)+psileft(1right)right)$$
                  $psi(1)=-gamma$ Euler-Mascheroni constant. And $psi(1/4)=-gamma-frac{pi}{2}-3ln(2)$
                  $$S_infty=frac12left(frac{pi}{2}+3ln(2)right)$$
                  $$sum _{r=1}^infty left(frac{2}{4r-3}-frac{1}{2r}right)=frac{pi}{4}+frac32ln(2)$$
                  For reference : http://mathworld.wolfram.com/DigammaFunction.html







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 29 '18 at 9:21









                  JJacquelinJJacquelin

                  42.8k21750




                  42.8k21750






























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                      張江高科駅

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                      張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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