Series sum with factorial notation Announcing the arrival of Valued Associate #679: Cesar...

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Series sum with factorial notation



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Sum of a simple infinite seriesSum of infinite series with Irrational termsOn a series involving harmonic numbersHow do I calculate the sum of this infinite series? $sumlimits_{n=1}^{infty}frac{1}{4n^{2}-1}$Finding $lim_{nrightarrowinfty} sum^{n}_{k=0}left|frac{2picos(kpi(3-sqrt{5}))}{n}right|$Convergence or not of infinite series: $sum^{infty}_{n=1}frac{n}{1+n^2}$Sum of series $frac{ 4}{10}+frac{4cdot 7}{10cdot 20}+frac{4cdot 7 cdot 10}{10cdot 20 cdot 30}+cdots cdots$Sum of $ frac{1}{1cdot 3}+frac{1}{1cdot 3cdot 5}+frac{1}{1cdot 3cdot 5cdot 7}+ cdots$Infinite series containing positive and negative termsInfinite series sum which have infinite terms












2












$begingroup$


The sum of series



$displaystyle 1+frac{1}{1!}cdot frac{1}{4}+frac{1cdot 3}{2!}cdot frac{1}{4^2}+frac{1cdot 3 cdot 5}{3!}cdot frac{1}{4^3}+cdots $



what i try:



$$1+lim_{nrightarrow infty}sum^{n}_{r=1}frac{1cdot 3cdot cdots (2r-1)}{r!cdot 4^r}$$



$$1+lim_{nrightarrow infty}sum^{n}_{r=1}frac{(2r)!}{r!prod^{n}_{r=1}(2r)!}$$



How do i solve it Help me please






sequences-and-series







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    what is the question...?
    $endgroup$
    – Jneven
    Mar 23 at 12:49
















2












$begingroup$


The sum of series



$displaystyle 1+frac{1}{1!}cdot frac{1}{4}+frac{1cdot 3}{2!}cdot frac{1}{4^2}+frac{1cdot 3 cdot 5}{3!}cdot frac{1}{4^3}+cdots $



what i try:



$$1+lim_{nrightarrow infty}sum^{n}_{r=1}frac{1cdot 3cdot cdots (2r-1)}{r!cdot 4^r}$$



$$1+lim_{nrightarrow infty}sum^{n}_{r=1}frac{(2r)!}{r!prod^{n}_{r=1}(2r)!}$$



How do i solve it Help me please






sequences-and-series







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    what is the question...?
    $endgroup$
    – Jneven
    Mar 23 at 12:49














2












2








2


1



$begingroup$


The sum of series



$displaystyle 1+frac{1}{1!}cdot frac{1}{4}+frac{1cdot 3}{2!}cdot frac{1}{4^2}+frac{1cdot 3 cdot 5}{3!}cdot frac{1}{4^3}+cdots $



what i try:



$$1+lim_{nrightarrow infty}sum^{n}_{r=1}frac{1cdot 3cdot cdots (2r-1)}{r!cdot 4^r}$$



$$1+lim_{nrightarrow infty}sum^{n}_{r=1}frac{(2r)!}{r!prod^{n}_{r=1}(2r)!}$$



How do i solve it Help me please






sequences-and-series







share|cite|improve this question









$endgroup$




The sum of series



$displaystyle 1+frac{1}{1!}cdot frac{1}{4}+frac{1cdot 3}{2!}cdot frac{1}{4^2}+frac{1cdot 3 cdot 5}{3!}cdot frac{1}{4^3}+cdots $



what i try:



$$1+lim_{nrightarrow infty}sum^{n}_{r=1}frac{1cdot 3cdot cdots (2r-1)}{r!cdot 4^r}$$



$$1+lim_{nrightarrow infty}sum^{n}_{r=1}frac{(2r)!}{r!prod^{n}_{r=1}(2r)!}$$



How do i solve it Help me please






sequences-and-series




sequences-and-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 23 at 11:59









jackyjacky

1,349816




1,349816








  • 1




    $begingroup$
    what is the question...?
    $endgroup$
    – Jneven
    Mar 23 at 12:49














  • 1




    $begingroup$
    what is the question...?
    $endgroup$
    – Jneven
    Mar 23 at 12:49








1




1




$begingroup$
what is the question...?
$endgroup$
– Jneven
Mar 23 at 12:49




$begingroup$
what is the question...?
$endgroup$
– Jneven
Mar 23 at 12:49










2 Answers
2






active

oldest

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3












$begingroup$


The series under consideration is



begin{align*}
color{blue}{1+sum_{r=1}^inftyfrac{(2r-1)!!}{r!4^r}}&=1+sum_{r=1}^inftyfrac{(2r)!}{r!4^r(2r)!!}\
&=1+sum_{r=1}^inftyfrac{(2r)!}{r!r!}left(frac{1}{8}right)^r\
&=sum_{r=0}^inftybinom{2r}{r}left(frac{1}{8}right)^r\
&=left.frac{1}{sqrt{1-4z}}right|_{z=1/8}tag{1}\
&=frac{1}{sqrt{1-frac{1}{2}}}\
&,,color{blue}{=sqrt{2}}
end{align*}




In (1) we use the ordinary generating function of the central binomial coefficients evaluated at $z=1/8$.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
    newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
    newcommand{dd}{mathrm{d}}
    newcommand{ds}[1]{displaystyle{#1}}
    newcommand{expo}[1]{,mathrm{e}^{#1},}
    newcommand{ic}{mathrm{i}}
    newcommand{mc}[1]{mathcal{#1}}
    newcommand{mrm}[1]{mathrm{#1}}
    newcommand{pars}[1]{left(,{#1},right)}
    newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
    newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
    newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
    newcommand{verts}[1]{leftvert,{#1},rightvert}$
    begin{align}
    &bbox[10px,#ffd]{1 + {1 over 1!},{1 over 4} +
    {1cdot 3 over 2!}, {1 over 4^{2}} +
    {1cdot 3 cdot 5 over 3!},{1 over 4^{3}} + cdots} =
    1 + sum_{n = 1}^{infty}{prod_{k = 0}^{n - 1}pars{2k + 1} over
    n!, 4^{n}}
    \[5mm] = &
    1 + sum_{n = 1}^{infty}{2^{n}prod_{k = 0}^{n - 1}pars{k + 1/2} over
    n!, 4^{n}} =
    1 + sum_{n = 1}^{infty}{pars{1/2}^{overline{n}} over n!}
    ,pars{1 over 2}^{n}
    \[5mm] = &
    1 + sum_{n = 1}^{infty}{Gammapars{1/2 + n}/Gammapars{1/2} over n!}
    ,pars{1 over 2}^{n} =
    sum_{n = 0}^{infty}{pars{n - 1/2}! over n!pars{-1/2}!}
    ,pars{1 over 2}^{n}
    \[5mm] = &
    sum_{n = 0}^{infty}{n - 1/2 choose n},pars{1 over 2}^{n} =
    sum_{n = 0}^{infty}{-1/2 choose n},pars{-,{1 over 2}}^{n} =
    bracks{1 + pars{-1 over 2}}^{-1/2}
    \[5mm] = & bbx{root{2}} approx 1.4142
    end{align}






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$


      The series under consideration is



      begin{align*}
      color{blue}{1+sum_{r=1}^inftyfrac{(2r-1)!!}{r!4^r}}&=1+sum_{r=1}^inftyfrac{(2r)!}{r!4^r(2r)!!}\
      &=1+sum_{r=1}^inftyfrac{(2r)!}{r!r!}left(frac{1}{8}right)^r\
      &=sum_{r=0}^inftybinom{2r}{r}left(frac{1}{8}right)^r\
      &=left.frac{1}{sqrt{1-4z}}right|_{z=1/8}tag{1}\
      &=frac{1}{sqrt{1-frac{1}{2}}}\
      &,,color{blue}{=sqrt{2}}
      end{align*}




      In (1) we use the ordinary generating function of the central binomial coefficients evaluated at $z=1/8$.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$


        The series under consideration is



        begin{align*}
        color{blue}{1+sum_{r=1}^inftyfrac{(2r-1)!!}{r!4^r}}&=1+sum_{r=1}^inftyfrac{(2r)!}{r!4^r(2r)!!}\
        &=1+sum_{r=1}^inftyfrac{(2r)!}{r!r!}left(frac{1}{8}right)^r\
        &=sum_{r=0}^inftybinom{2r}{r}left(frac{1}{8}right)^r\
        &=left.frac{1}{sqrt{1-4z}}right|_{z=1/8}tag{1}\
        &=frac{1}{sqrt{1-frac{1}{2}}}\
        &,,color{blue}{=sqrt{2}}
        end{align*}




        In (1) we use the ordinary generating function of the central binomial coefficients evaluated at $z=1/8$.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$


          The series under consideration is



          begin{align*}
          color{blue}{1+sum_{r=1}^inftyfrac{(2r-1)!!}{r!4^r}}&=1+sum_{r=1}^inftyfrac{(2r)!}{r!4^r(2r)!!}\
          &=1+sum_{r=1}^inftyfrac{(2r)!}{r!r!}left(frac{1}{8}right)^r\
          &=sum_{r=0}^inftybinom{2r}{r}left(frac{1}{8}right)^r\
          &=left.frac{1}{sqrt{1-4z}}right|_{z=1/8}tag{1}\
          &=frac{1}{sqrt{1-frac{1}{2}}}\
          &,,color{blue}{=sqrt{2}}
          end{align*}




          In (1) we use the ordinary generating function of the central binomial coefficients evaluated at $z=1/8$.






          share|cite|improve this answer











          $endgroup$




          The series under consideration is



          begin{align*}
          color{blue}{1+sum_{r=1}^inftyfrac{(2r-1)!!}{r!4^r}}&=1+sum_{r=1}^inftyfrac{(2r)!}{r!4^r(2r)!!}\
          &=1+sum_{r=1}^inftyfrac{(2r)!}{r!r!}left(frac{1}{8}right)^r\
          &=sum_{r=0}^inftybinom{2r}{r}left(frac{1}{8}right)^r\
          &=left.frac{1}{sqrt{1-4z}}right|_{z=1/8}tag{1}\
          &=frac{1}{sqrt{1-frac{1}{2}}}\
          &,,color{blue}{=sqrt{2}}
          end{align*}




          In (1) we use the ordinary generating function of the central binomial coefficients evaluated at $z=1/8$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 23 at 16:13

























          answered Mar 23 at 15:59









          Markus ScheuerMarkus Scheuer

          64.4k460152




          64.4k460152























              0












              $begingroup$

              $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
              newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
              newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
              newcommand{dd}{mathrm{d}}
              newcommand{ds}[1]{displaystyle{#1}}
              newcommand{expo}[1]{,mathrm{e}^{#1},}
              newcommand{ic}{mathrm{i}}
              newcommand{mc}[1]{mathcal{#1}}
              newcommand{mrm}[1]{mathrm{#1}}
              newcommand{pars}[1]{left(,{#1},right)}
              newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
              newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
              newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
              newcommand{verts}[1]{leftvert,{#1},rightvert}$
              begin{align}
              &bbox[10px,#ffd]{1 + {1 over 1!},{1 over 4} +
              {1cdot 3 over 2!}, {1 over 4^{2}} +
              {1cdot 3 cdot 5 over 3!},{1 over 4^{3}} + cdots} =
              1 + sum_{n = 1}^{infty}{prod_{k = 0}^{n - 1}pars{2k + 1} over
              n!, 4^{n}}
              \[5mm] = &
              1 + sum_{n = 1}^{infty}{2^{n}prod_{k = 0}^{n - 1}pars{k + 1/2} over
              n!, 4^{n}} =
              1 + sum_{n = 1}^{infty}{pars{1/2}^{overline{n}} over n!}
              ,pars{1 over 2}^{n}
              \[5mm] = &
              1 + sum_{n = 1}^{infty}{Gammapars{1/2 + n}/Gammapars{1/2} over n!}
              ,pars{1 over 2}^{n} =
              sum_{n = 0}^{infty}{pars{n - 1/2}! over n!pars{-1/2}!}
              ,pars{1 over 2}^{n}
              \[5mm] = &
              sum_{n = 0}^{infty}{n - 1/2 choose n},pars{1 over 2}^{n} =
              sum_{n = 0}^{infty}{-1/2 choose n},pars{-,{1 over 2}}^{n} =
              bracks{1 + pars{-1 over 2}}^{-1/2}
              \[5mm] = & bbx{root{2}} approx 1.4142
              end{align}






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                newcommand{dd}{mathrm{d}}
                newcommand{ds}[1]{displaystyle{#1}}
                newcommand{expo}[1]{,mathrm{e}^{#1},}
                newcommand{ic}{mathrm{i}}
                newcommand{mc}[1]{mathcal{#1}}
                newcommand{mrm}[1]{mathrm{#1}}
                newcommand{pars}[1]{left(,{#1},right)}
                newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
                newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
                newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                newcommand{verts}[1]{leftvert,{#1},rightvert}$
                begin{align}
                &bbox[10px,#ffd]{1 + {1 over 1!},{1 over 4} +
                {1cdot 3 over 2!}, {1 over 4^{2}} +
                {1cdot 3 cdot 5 over 3!},{1 over 4^{3}} + cdots} =
                1 + sum_{n = 1}^{infty}{prod_{k = 0}^{n - 1}pars{2k + 1} over
                n!, 4^{n}}
                \[5mm] = &
                1 + sum_{n = 1}^{infty}{2^{n}prod_{k = 0}^{n - 1}pars{k + 1/2} over
                n!, 4^{n}} =
                1 + sum_{n = 1}^{infty}{pars{1/2}^{overline{n}} over n!}
                ,pars{1 over 2}^{n}
                \[5mm] = &
                1 + sum_{n = 1}^{infty}{Gammapars{1/2 + n}/Gammapars{1/2} over n!}
                ,pars{1 over 2}^{n} =
                sum_{n = 0}^{infty}{pars{n - 1/2}! over n!pars{-1/2}!}
                ,pars{1 over 2}^{n}
                \[5mm] = &
                sum_{n = 0}^{infty}{n - 1/2 choose n},pars{1 over 2}^{n} =
                sum_{n = 0}^{infty}{-1/2 choose n},pars{-,{1 over 2}}^{n} =
                bracks{1 + pars{-1 over 2}}^{-1/2}
                \[5mm] = & bbx{root{2}} approx 1.4142
                end{align}






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                  newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                  newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                  newcommand{dd}{mathrm{d}}
                  newcommand{ds}[1]{displaystyle{#1}}
                  newcommand{expo}[1]{,mathrm{e}^{#1},}
                  newcommand{ic}{mathrm{i}}
                  newcommand{mc}[1]{mathcal{#1}}
                  newcommand{mrm}[1]{mathrm{#1}}
                  newcommand{pars}[1]{left(,{#1},right)}
                  newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
                  newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
                  newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                  newcommand{verts}[1]{leftvert,{#1},rightvert}$
                  begin{align}
                  &bbox[10px,#ffd]{1 + {1 over 1!},{1 over 4} +
                  {1cdot 3 over 2!}, {1 over 4^{2}} +
                  {1cdot 3 cdot 5 over 3!},{1 over 4^{3}} + cdots} =
                  1 + sum_{n = 1}^{infty}{prod_{k = 0}^{n - 1}pars{2k + 1} over
                  n!, 4^{n}}
                  \[5mm] = &
                  1 + sum_{n = 1}^{infty}{2^{n}prod_{k = 0}^{n - 1}pars{k + 1/2} over
                  n!, 4^{n}} =
                  1 + sum_{n = 1}^{infty}{pars{1/2}^{overline{n}} over n!}
                  ,pars{1 over 2}^{n}
                  \[5mm] = &
                  1 + sum_{n = 1}^{infty}{Gammapars{1/2 + n}/Gammapars{1/2} over n!}
                  ,pars{1 over 2}^{n} =
                  sum_{n = 0}^{infty}{pars{n - 1/2}! over n!pars{-1/2}!}
                  ,pars{1 over 2}^{n}
                  \[5mm] = &
                  sum_{n = 0}^{infty}{n - 1/2 choose n},pars{1 over 2}^{n} =
                  sum_{n = 0}^{infty}{-1/2 choose n},pars{-,{1 over 2}}^{n} =
                  bracks{1 + pars{-1 over 2}}^{-1/2}
                  \[5mm] = & bbx{root{2}} approx 1.4142
                  end{align}






                  share|cite|improve this answer









                  $endgroup$



                  $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                  newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                  newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                  newcommand{dd}{mathrm{d}}
                  newcommand{ds}[1]{displaystyle{#1}}
                  newcommand{expo}[1]{,mathrm{e}^{#1},}
                  newcommand{ic}{mathrm{i}}
                  newcommand{mc}[1]{mathcal{#1}}
                  newcommand{mrm}[1]{mathrm{#1}}
                  newcommand{pars}[1]{left(,{#1},right)}
                  newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
                  newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
                  newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                  newcommand{verts}[1]{leftvert,{#1},rightvert}$
                  begin{align}
                  &bbox[10px,#ffd]{1 + {1 over 1!},{1 over 4} +
                  {1cdot 3 over 2!}, {1 over 4^{2}} +
                  {1cdot 3 cdot 5 over 3!},{1 over 4^{3}} + cdots} =
                  1 + sum_{n = 1}^{infty}{prod_{k = 0}^{n - 1}pars{2k + 1} over
                  n!, 4^{n}}
                  \[5mm] = &
                  1 + sum_{n = 1}^{infty}{2^{n}prod_{k = 0}^{n - 1}pars{k + 1/2} over
                  n!, 4^{n}} =
                  1 + sum_{n = 1}^{infty}{pars{1/2}^{overline{n}} over n!}
                  ,pars{1 over 2}^{n}
                  \[5mm] = &
                  1 + sum_{n = 1}^{infty}{Gammapars{1/2 + n}/Gammapars{1/2} over n!}
                  ,pars{1 over 2}^{n} =
                  sum_{n = 0}^{infty}{pars{n - 1/2}! over n!pars{-1/2}!}
                  ,pars{1 over 2}^{n}
                  \[5mm] = &
                  sum_{n = 0}^{infty}{n - 1/2 choose n},pars{1 over 2}^{n} =
                  sum_{n = 0}^{infty}{-1/2 choose n},pars{-,{1 over 2}}^{n} =
                  bracks{1 + pars{-1 over 2}}^{-1/2}
                  \[5mm] = & bbx{root{2}} approx 1.4142
                  end{align}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 27 at 20:24









                  Felix MarinFelix Marin

                  68.9k7110147




                  68.9k7110147






























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                      Fil:Tokke komm.svg

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                      Skildring Skildring Tokke komm.svg Norsk nynorsk: Våpen for Tokke kommune i Telemark fylke, godkjent ved kongeleg resolusjon 6. februar 1987. Motiv: På gull grunn ein gåande svart bjørn. [1] Tokke ligg i eit gamalt bjørneområde og bjørnen er eit viktig innslag i fleire lokale historier. Bjørnen kan òg symbolisere naturen og dyrelivet i kommunen. English: Coat of arms of the Norwegian municipality of Tokke, Telemark. Dato 6. februar 1987 Kjelde Original by Stein Davidsen, based upon image from caplex.no. Opphavsperson Eige arbeid Vektorgrafikken vart laga med Inkscape. Løyve (Gjenbruk av denne fila) Public domain Public domain false false This official Norwegian coat of arms is ineligible for copyright, but its usage is restricted according to Norwegian penal law, independent of the copyright status of the depiction shown here. Dette biletet er sett saman av eit flagg, eit våpen, eit sigill eller eit anna offisielt insignium . Bruken av slike symbol er avgrensa i fleire land. Desse r
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