How does this infinite series simplify to an integral? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Infinite Series ManipulationsHow do I solve this integral? (looking for hints)Finding Sum for Infinite SeriesSimplify Infinite Series Involving Gamma Function $Gamma$How to simplify this integral?Does this infinite series converge or diverge?Sum of this infinite series…Why does Improper Integral Diverge?Infinite series containing positive and negative termsInfinite series sum which have infinite terms

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How does this infinite series simplify to an integral?

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How does this infinite series simplify to an integral?



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Infinite Series ManipulationsHow do I solve this integral? (looking for hints)Finding Sum for Infinite SeriesSimplify Infinite Series Involving Gamma Function $Gamma$How to simplify this integral?Does this infinite series converge or diverge?Sum of this infinite series…Why does Improper Integral Diverge?Infinite series containing positive and negative termsInfinite series sum which have infinite terms










2












$begingroup$


How does the infinite series in the link above simplify to that integral?



$$1-frac14+frac17-frac110+cdots=int_0^1fracdx1+x^3$$



I thought of simplifying the series to the sum to infinity of $frac16n-5 - frac16n-2$, but this did not help.






integration sequences-and-series power-series improper-integrals







share|cite











$endgroup$







  • 1




    $begingroup$
    en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
    $endgroup$
    – lab bhattacharjee
    4 hours ago










  • $begingroup$
    It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
    $endgroup$
    – StubbornAtom
    4 hours ago















2












$begingroup$


How does the infinite series in the link above simplify to that integral?



$$1-frac14+frac17-frac110+cdots=int_0^1fracdx1+x^3$$



I thought of simplifying the series to the sum to infinity of $frac16n-5 - frac16n-2$, but this did not help.






integration sequences-and-series power-series improper-integrals







share|cite











$endgroup$







  • 1




    $begingroup$
    en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
    $endgroup$
    – lab bhattacharjee
    4 hours ago










  • $begingroup$
    It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
    $endgroup$
    – StubbornAtom
    4 hours ago













2












2








2


0



$begingroup$


How does the infinite series in the link above simplify to that integral?



$$1-frac14+frac17-frac110+cdots=int_0^1fracdx1+x^3$$



I thought of simplifying the series to the sum to infinity of $frac16n-5 - frac16n-2$, but this did not help.






integration sequences-and-series power-series improper-integrals







share|cite











$endgroup$




How does the infinite series in the link above simplify to that integral?



$$1-frac14+frac17-frac110+cdots=int_0^1fracdx1+x^3$$



I thought of simplifying the series to the sum to infinity of $frac16n-5 - frac16n-2$, but this did not help.






integration sequences-and-series power-series improper-integrals




integration sequences-and-series power-series improper-integrals






share|cite















share|cite













share|cite




share|cite








edited 4 hours ago









Zacky

7,87511062




7,87511062










asked 4 hours ago









ShreeShree

114




114







  • 1




    $begingroup$
    en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
    $endgroup$
    – lab bhattacharjee
    4 hours ago










  • $begingroup$
    It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
    $endgroup$
    – StubbornAtom
    4 hours ago












  • 1




    $begingroup$
    en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
    $endgroup$
    – lab bhattacharjee
    4 hours ago










  • $begingroup$
    It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
    $endgroup$
    – StubbornAtom
    4 hours ago







1




1




$begingroup$
en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
$endgroup$
– lab bhattacharjee
4 hours ago




$begingroup$
en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
$endgroup$
– lab bhattacharjee
4 hours ago












$begingroup$
It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
$endgroup$
– StubbornAtom
4 hours ago




$begingroup$
It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
$endgroup$
– StubbornAtom
4 hours ago










3 Answers
3






active

oldest

votes


















6












$begingroup$

$$int_0^1fracdx1-(-x)^3=int_0^1sum_n=0^infty(-x)^3ndx=sum_n=0^infty(-1)^3nint_0^1x^3ndx$$
$$=sum_n=0^inftyfrac(-1)^3n3n+1= 1-frac14+frac17-frac110+cdots $$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    For those who aren't clear on the first equality, it's a geometric series formula.
    $endgroup$
    – Alexis Olson
    1 hour ago


















2












$begingroup$

for $x$ real, $ngeq 0$ integer
beginalignfrac11+x^3&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11-(-x^3)\
&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11+x^3\
endalign



For $xneq 1$, $ngeq 0$ integer, beginalignsum_k=0^n x^k=frac1-x^n+11-xendalign



Therefore,
beginalignint_0^1 frac11+x^3,dx&=int_0^1 left(sum_k=0^n (-x^3)^kright),dx+int_0^1 frac(-x^3)^n+11+x^3,dx\
&=sum_k=0^n left(int_0^1 (-x^3)^k,dxright)+int_0^1 frac(-x^3)^n+11+x^3,dx\
&=sum_k=0^n frac(-1)^k3k+1+int_0^1 frac(-x^3)^n+11+x^3,dx\
endalign



For $xin[0;1],ngeq 0$, integer,
beginalignfracx^3(n+1)1+x^3leq x^3(n+1)endalign
and,
beginalignint_0^1 x^3(n+1),dx=frac13n+4endalign
Therefore,
beginalignleft|int_0^1 frac(-x^3)^n+11+x^3,dxright|leq frac13n+4endalign
beginalignleft|int_0^1 frac11+x^3,dx-sum_k=0^n frac(-1)^k3k+1right|leq frac13n+4endalign
Therefore,
beginalignboxedint_0^1 frac11+x^3,dx=sum_k=0^infty frac(-1)^k3k+1endalign






share|cite|improve this answer









$endgroup$




















    1












    $begingroup$

    If $lvert xrvert<1$, let$$f(x)=sum_n=0^inftyfracx^3n+13n+1.$$Then $$f'(x)=sum_n=0^infty x^3n=frac11-x^3.$$Thereforebeginalign1-frac14+frac17-frac110+cdots&=lim_xto1f(x)\&=int_0^1f'(x),mathrm dx\&=int_0^1frac11-x^3,mathrm dx.endalign






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      $$int_0^1fracdx1-(-x)^3=int_0^1sum_n=0^infty(-x)^3ndx=sum_n=0^infty(-1)^3nint_0^1x^3ndx$$
      $$=sum_n=0^inftyfrac(-1)^3n3n+1= 1-frac14+frac17-frac110+cdots $$






      share|cite|improve this answer











      $endgroup$












      • $begingroup$
        For those who aren't clear on the first equality, it's a geometric series formula.
        $endgroup$
        – Alexis Olson
        1 hour ago















      6












      $begingroup$

      $$int_0^1fracdx1-(-x)^3=int_0^1sum_n=0^infty(-x)^3ndx=sum_n=0^infty(-1)^3nint_0^1x^3ndx$$
      $$=sum_n=0^inftyfrac(-1)^3n3n+1= 1-frac14+frac17-frac110+cdots $$






      share|cite|improve this answer











      $endgroup$












      • $begingroup$
        For those who aren't clear on the first equality, it's a geometric series formula.
        $endgroup$
        – Alexis Olson
        1 hour ago













      6












      6








      6





      $begingroup$

      $$int_0^1fracdx1-(-x)^3=int_0^1sum_n=0^infty(-x)^3ndx=sum_n=0^infty(-1)^3nint_0^1x^3ndx$$
      $$=sum_n=0^inftyfrac(-1)^3n3n+1= 1-frac14+frac17-frac110+cdots $$






      share|cite|improve this answer











      $endgroup$



      $$int_0^1fracdx1-(-x)^3=int_0^1sum_n=0^infty(-x)^3ndx=sum_n=0^infty(-1)^3nint_0^1x^3ndx$$
      $$=sum_n=0^inftyfrac(-1)^3n3n+1= 1-frac14+frac17-frac110+cdots $$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 18 mins ago

























      answered 4 hours ago









      HAMIDINE SOUMAREHAMIDINE SOUMARE

      2,194212




      2,194212











      • $begingroup$
        For those who aren't clear on the first equality, it's a geometric series formula.
        $endgroup$
        – Alexis Olson
        1 hour ago
















      • $begingroup$
        For those who aren't clear on the first equality, it's a geometric series formula.
        $endgroup$
        – Alexis Olson
        1 hour ago















      $begingroup$
      For those who aren't clear on the first equality, it's a geometric series formula.
      $endgroup$
      – Alexis Olson
      1 hour ago




      $begingroup$
      For those who aren't clear on the first equality, it's a geometric series formula.
      $endgroup$
      – Alexis Olson
      1 hour ago











      2












      $begingroup$

      for $x$ real, $ngeq 0$ integer
      beginalignfrac11+x^3&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11-(-x^3)\
      &=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11+x^3\
      endalign



      For $xneq 1$, $ngeq 0$ integer, beginalignsum_k=0^n x^k=frac1-x^n+11-xendalign



      Therefore,
      beginalignint_0^1 frac11+x^3,dx&=int_0^1 left(sum_k=0^n (-x^3)^kright),dx+int_0^1 frac(-x^3)^n+11+x^3,dx\
      &=sum_k=0^n left(int_0^1 (-x^3)^k,dxright)+int_0^1 frac(-x^3)^n+11+x^3,dx\
      &=sum_k=0^n frac(-1)^k3k+1+int_0^1 frac(-x^3)^n+11+x^3,dx\
      endalign



      For $xin[0;1],ngeq 0$, integer,
      beginalignfracx^3(n+1)1+x^3leq x^3(n+1)endalign
      and,
      beginalignint_0^1 x^3(n+1),dx=frac13n+4endalign
      Therefore,
      beginalignleft|int_0^1 frac(-x^3)^n+11+x^3,dxright|leq frac13n+4endalign
      beginalignleft|int_0^1 frac11+x^3,dx-sum_k=0^n frac(-1)^k3k+1right|leq frac13n+4endalign
      Therefore,
      beginalignboxedint_0^1 frac11+x^3,dx=sum_k=0^infty frac(-1)^k3k+1endalign






      share|cite|improve this answer









      $endgroup$

















        2












        $begingroup$

        for $x$ real, $ngeq 0$ integer
        beginalignfrac11+x^3&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11-(-x^3)\
        &=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11+x^3\
        endalign



        For $xneq 1$, $ngeq 0$ integer, beginalignsum_k=0^n x^k=frac1-x^n+11-xendalign



        Therefore,
        beginalignint_0^1 frac11+x^3,dx&=int_0^1 left(sum_k=0^n (-x^3)^kright),dx+int_0^1 frac(-x^3)^n+11+x^3,dx\
        &=sum_k=0^n left(int_0^1 (-x^3)^k,dxright)+int_0^1 frac(-x^3)^n+11+x^3,dx\
        &=sum_k=0^n frac(-1)^k3k+1+int_0^1 frac(-x^3)^n+11+x^3,dx\
        endalign



        For $xin[0;1],ngeq 0$, integer,
        beginalignfracx^3(n+1)1+x^3leq x^3(n+1)endalign
        and,
        beginalignint_0^1 x^3(n+1),dx=frac13n+4endalign
        Therefore,
        beginalignleft|int_0^1 frac(-x^3)^n+11+x^3,dxright|leq frac13n+4endalign
        beginalignleft|int_0^1 frac11+x^3,dx-sum_k=0^n frac(-1)^k3k+1right|leq frac13n+4endalign
        Therefore,
        beginalignboxedint_0^1 frac11+x^3,dx=sum_k=0^infty frac(-1)^k3k+1endalign






        share|cite|improve this answer









        $endgroup$















          2












          2








          2





          $begingroup$

          for $x$ real, $ngeq 0$ integer
          beginalignfrac11+x^3&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11-(-x^3)\
          &=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11+x^3\
          endalign



          For $xneq 1$, $ngeq 0$ integer, beginalignsum_k=0^n x^k=frac1-x^n+11-xendalign



          Therefore,
          beginalignint_0^1 frac11+x^3,dx&=int_0^1 left(sum_k=0^n (-x^3)^kright),dx+int_0^1 frac(-x^3)^n+11+x^3,dx\
          &=sum_k=0^n left(int_0^1 (-x^3)^k,dxright)+int_0^1 frac(-x^3)^n+11+x^3,dx\
          &=sum_k=0^n frac(-1)^k3k+1+int_0^1 frac(-x^3)^n+11+x^3,dx\
          endalign



          For $xin[0;1],ngeq 0$, integer,
          beginalignfracx^3(n+1)1+x^3leq x^3(n+1)endalign
          and,
          beginalignint_0^1 x^3(n+1),dx=frac13n+4endalign
          Therefore,
          beginalignleft|int_0^1 frac(-x^3)^n+11+x^3,dxright|leq frac13n+4endalign
          beginalignleft|int_0^1 frac11+x^3,dx-sum_k=0^n frac(-1)^k3k+1right|leq frac13n+4endalign
          Therefore,
          beginalignboxedint_0^1 frac11+x^3,dx=sum_k=0^infty frac(-1)^k3k+1endalign






          share|cite|improve this answer









          $endgroup$



          for $x$ real, $ngeq 0$ integer
          beginalignfrac11+x^3&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11-(-x^3)\
          &=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11+x^3\
          endalign



          For $xneq 1$, $ngeq 0$ integer, beginalignsum_k=0^n x^k=frac1-x^n+11-xendalign



          Therefore,
          beginalignint_0^1 frac11+x^3,dx&=int_0^1 left(sum_k=0^n (-x^3)^kright),dx+int_0^1 frac(-x^3)^n+11+x^3,dx\
          &=sum_k=0^n left(int_0^1 (-x^3)^k,dxright)+int_0^1 frac(-x^3)^n+11+x^3,dx\
          &=sum_k=0^n frac(-1)^k3k+1+int_0^1 frac(-x^3)^n+11+x^3,dx\
          endalign



          For $xin[0;1],ngeq 0$, integer,
          beginalignfracx^3(n+1)1+x^3leq x^3(n+1)endalign
          and,
          beginalignint_0^1 x^3(n+1),dx=frac13n+4endalign
          Therefore,
          beginalignleft|int_0^1 frac(-x^3)^n+11+x^3,dxright|leq frac13n+4endalign
          beginalignleft|int_0^1 frac11+x^3,dx-sum_k=0^n frac(-1)^k3k+1right|leq frac13n+4endalign
          Therefore,
          beginalignboxedint_0^1 frac11+x^3,dx=sum_k=0^infty frac(-1)^k3k+1endalign







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          FDPFDP

          6,12211929




          6,12211929





















              1












              $begingroup$

              If $lvert xrvert<1$, let$$f(x)=sum_n=0^inftyfracx^3n+13n+1.$$Then $$f'(x)=sum_n=0^infty x^3n=frac11-x^3.$$Thereforebeginalign1-frac14+frac17-frac110+cdots&=lim_xto1f(x)\&=int_0^1f'(x),mathrm dx\&=int_0^1frac11-x^3,mathrm dx.endalign






              share|cite|improve this answer









              $endgroup$

















                1












                $begingroup$

                If $lvert xrvert<1$, let$$f(x)=sum_n=0^inftyfracx^3n+13n+1.$$Then $$f'(x)=sum_n=0^infty x^3n=frac11-x^3.$$Thereforebeginalign1-frac14+frac17-frac110+cdots&=lim_xto1f(x)\&=int_0^1f'(x),mathrm dx\&=int_0^1frac11-x^3,mathrm dx.endalign






                share|cite|improve this answer









                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  If $lvert xrvert<1$, let$$f(x)=sum_n=0^inftyfracx^3n+13n+1.$$Then $$f'(x)=sum_n=0^infty x^3n=frac11-x^3.$$Thereforebeginalign1-frac14+frac17-frac110+cdots&=lim_xto1f(x)\&=int_0^1f'(x),mathrm dx\&=int_0^1frac11-x^3,mathrm dx.endalign






                  share|cite|improve this answer









                  $endgroup$



                  If $lvert xrvert<1$, let$$f(x)=sum_n=0^inftyfracx^3n+13n+1.$$Then $$f'(x)=sum_n=0^infty x^3n=frac11-x^3.$$Thereforebeginalign1-frac14+frac17-frac110+cdots&=lim_xto1f(x)\&=int_0^1f'(x),mathrm dx\&=int_0^1frac11-x^3,mathrm dx.endalign







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 4 hours ago









                  José Carlos SantosJosé Carlos Santos

                  174k23133243




                  174k23133243



























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                      Luksemburg Sisukord Nimi | Asend | Loodus | Riigikord | Haldusjaotus | Rahvastik | Riigikaitse | Majandus | Taristu | Ajalugu | Eesti ja Luksemburgi suhted | Haridus | Kultuur | Vaata ka | Viited | Välislingid | Navigeerimismenüü50° N, 6° EÜlevaade Luksemburgi kaitsealadest.Luksemburgi rahvaarv. Statistikaamet.World Bank'i andmebaasÜlevaade Luksemburgi loodusest.Ülevaade Luksemburgi metsadest.Guy Colling. "Red List of the Vascular Plants of Luxembourg." Travaux scientifiques du Musée national d’histoire naturelle Luxembourg. 2005.Luxembourg’s biodiversity at risk.Maailma kahepaiksete andmebaas.Denis Lepage. "Luxembourg." Avibase.Ülevaade temperatuuridest. Luksemburgi meteoroloogiateenistus.Ülevaade Luksemburgist. Euroopa Liidu esinduse koduleht.Système politique. TerritoireÜlevaade Luksemburgi rahvastikust. Luksemburgi statistikaamet.Luksemburgi rahvastik. Luksemburgi statistikaamet.The World FactbookMonique Borsenberger, Paul Dickes. "Religions au Luxembourg. Quelle évolution entre 1999-2008". Luksemburgi statistikaamet. 2011.Luksemburgi peapiiskopkond. Catholic-Hierarchy.Luksemburgi armee koduleht.Luksemburgi armee relvastus.Eesti Välisministeerium.Luksemburgi rahvastik. Luksemburgi statistikaamet.Luksemburgi Eesti Seltsi koduleht.Helen Eelrand. "Raadio, mis muutis maailma." Eesti Päevaleht. 13. märts 2004.Ülevaade Luksemburgi haridussüsteemist.Ülevaade Luksemburgi keskkoolidest.Luksemburgr

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                      RiigidEuroopa maadLuksemburg merepiiritariikEuroopasMadalmaadeHollandiBelgiagaBeneluxiPrantsusmaaSaksamaaLuxembourgikindlusestSaksa keeles963SiegfriedileläänestpõhjastidastRheinland-PfalziSaarimaalõunastLorraineMoselleMeurthe-et-Moselle'i departemangudArdennideMoseli jõejõedmetsagaharilik pöökharilik tammhall nelkharilik kortslehtmageveekalukahepaikseidpäriskonnalisiroomajaidimetajaidsaarmasNatura 20001964EuroopasÜlevaade Luksemburgi kaitsealadest.parasvöötme kliimasuvejaanuarveebruarjuuliaugustkonstitutsiooniline monarhiaTäidesaatvat võimuSeadusandlik võim2013põhiseadus18421868NATOEuroopa Liidu1. jaanuari20171985letseburgi keelprantsuse keelkodakondsusePortugaliPrantsusmaaItaaliakatoliiklust187019882013ilmikpreestritajateenistusBosnia ja Hertsegoviinaskindral2000politseijõudLiechtensteiniKataritöötusSaksamaaBelgiaPrantsusmaaHollandmineraalidHiinaeurotsooniga1999euro2002sadamKeldi2. sajandil e...
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                      Valle di Casies Indice Geografia fisica | Origini del nome | Storia | Società | Amministrazione | Sport | Note | Bibliografia | Voci correlate | Altri progetti | Collegamenti esterni | Menu di navigazione46°46′N 12°11′E / 46.766667°N 12.183333°E46.766667; 12.183333 (Valle di Casies)46°46′N 12°11′E / 46.766667°N 12.183333°E46.766667; 12.183333 (Valle di Casies)Sito istituzionaleAstat Censimento della popolazione 2011 - Determinazione della consistenza dei tre gruppi linguistici della Provincia Autonoma di Bolzano-Alto Adige - giugno 2012Numeri e fattiValle di CasiesDato IstatTabella dei gradi/giorno dei Comuni italiani raggruppati per Regione e Provincia26 agosto 1993, n. 412Heraldry of the World: GsiesStatistiche I.StatValCasies.comWikimedia CommonsWikimedia CommonsValle di CasiesSito ufficialeValle di CasiesMM14870458910042978-6

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                      Valle di Casies tedescocomune italianoprovincia autonoma di BolzanoTrentino-Alto AdigeValVal Pusteriario CasiesMonguelfoAustriatoponimo117811891299tedescoXII secolo1929municipio1985ted.cappuccinoJoachim HaspingerAndreas HoferbavaresifrancesinapoleonicibronzoOthmar Winklerbunkersbarramento Planca di SopraVallo Alpino in Alto Adigesci di fondoGran fondo della Val Casiesskiliftpiste da slittino naturale Valle di Casies Da Wikipedia, l'enciclopedia libera. Jump to navigation Jump to search Valle di Casies comune ( IT ) Valle di Casies ( DE ) Gsies Una veduta della vallata Localizzazione Stato   Italia Regione Trentino-Alto Adige Provincia Bolzano Amministrazione Sindaco Kurti Taschler (SVP) dal 10-5-2015 (1º mandato) Lingue ufficiali Tedesco, Italiano Territorio Coordinate 46°46′N 12°11′E  /  46.766667°N 12.183333°E 46.766667; 12.183333  ( Valle di Casies ) Coordinate: 46°46...
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