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

What is the largest species of polychaete?

How do I keep my slimes from escaping their pens?

Do working physicists consider Newtonian mechanics to be "falsified"?

How should I respond to a player wanting to catch a sword between their hands?

How do I automatically answer y in bash script?

How to stop my camera from exagerrating differences in skin colour?

How is simplicity better than precision and clarity in prose?

Biased dice probability question

I'm thinking of a number

Estimated State payment too big --> money back; + 2018 Tax Reform

Should you tell Jews they are breaking a commandment?

Classification of bundles, Postnikov towers, obstruction theory, local coefficients

Single author papers against my advisor's will?

Can a monk deflect thrown melee weapons?

I'm having difficulty getting my players to do stuff in a sandbox campaign

Fishing simulator

Unexpected result with right shift after bitwise negation

Complexity of many constant time steps with occasional logarithmic steps

Why is there no army of Iron-Mans in the MCU?

Area of a 2D convex hull

Jazz greats knew nothing of modes. Why are they used to improvise on standards?

If A makes B more likely then B makes A more likely"

New Order #5: where Fibonacci and Beatty meet at Wythoff

Stars Make Stars



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

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











$endgroup$





















    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









    $endgroup$














      Your Answer








      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3159232%2fseries-sum-with-factorial-notation%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      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






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3159232%2fseries-sum-with-factorial-notation%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Nidaros erkebispedøme

                      Image
                      For det lutherske bispedømet, sjå Nidaros bispedøme. Nidaros erkebispedøme var eit katolsk erkebispedøme med sete i Nidaros. Det vart oppretta av kong Olav Kyrre mellom 1070 og 1080 som bispedøme. Bispedømet vart opphøgd til erkebispedøme i 1153. Erkebispedømet vart innstifta av pave Anastasius IV i 1154, og var i funksjon fram til reformasjonen i 1537. Erkebiskopen sitt hovudsete var Erkebispegarden, rett ved domkyrkja Nidarosdomen. Frå 1152 til 1154 var Nicholas Breakspear i Skandinavia som paveleg utsending, og organiserte opprettinga av erkesetet. Innhaldsliste 1 Utbreiing og jurisdiksjon 2 Overhyrdar 2.1 Lydbiskopar 3 Biskopar 3.1 Biskopar 3.2 Erkebiskopar Utbreiing og jurisdiksjon | Kart over den nordiske kyrkjeprovinsen (1153-1387) Erkebispedømet omfatta i tillegg til Noreg øyane i vest i Noregsveldet. Nidaros var eit metropolitanerkebispedøme med fylgjande underliggjande bispedøme suffraganar: Bj...
                      Read more

                      Birsay

                      Image
                      Tettstaden Birsay på Orknøyane. Ruinane av jarleplasset om lag midt i biletet. (Fotografi frå 2012) Utsikt mot Birsay og Brough of Brisay (fotografi frå 2009) Birsay er ein tettstad i eit sokn (parish) med same namn, nordvest på øya Mainland [Note 1] på Orknøyane, om lag 31 km frå Kirkwall. Staden er omtala fleire stader i Orknøyingasoga [1] , der han er kalla «Byrgesherad» (norrønt: Birgisherað ). Namnet skriv seg truleg frå elvenamnet Birgisá [2] . Innhaldsliste 1 Den gamle hovudstaden 2 Bispesetet 3 Jarlepalasset 4 Kjelder og utfyllande opplysningar 4.1 Notar 4.2 Fotnotar 4.3 Bakgrunnsstoff Den gamle hovudstaden | Hovudartikkelen for dette emnet er Torfinn Sigurdsson . Soknet er i dag i hovudsak ei jordbrukskommune men er mest kjend for dei mange kulturminna i området. Tettstaden Birsay blir gjerne omtala som «den gamle hovudstaden på Orknøyane» [3] , og ein viser d...
                      Read more

                      Was Woodrow Wilson really a Liberal?Was World War I a war of liberals against authoritarians?Founding Fathers...

                      Image
                      Do VLANs within a subnet need to have their own subnet for router on a stick? What are the differences between the usage of 'it' and 'they'? Dragon forelimb placement How do we improve the relationship with a client software team that performs poorly and is becoming less collaborative? What typically incentivizes a professor to change jobs to a lower ranking university? How to write a macro that is braces sensitive? Modeling an IPv4 Address Writing rule stating superpower from different root cause is bad writing Can I ask the recruiters in my resume to put the reason why I am rejected? Approximately how much travel t...
                      Read more