$lim_nrightarrow infty ( arctanfrac12 + arctan frac12.2^2 +…+ arctan frac 12n^2)$

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
2
down vote

favorite
2













Calculate
$$lim_nrightarrow infty left( arctanfrac12 + arctan frac12.2^2 +....+ arctan frac 12n^2right)$$




My answer: i know that $$ sum_n=1^N arctan left( frac2n^2 right) =sum_n=1^N arctan (n+1)-arctan(n-1)$$



as Im not able To find the $sum_k=1^n arctan frac 12k^2$



I need help,,,,,any hints /solution will be aprreciated



thanks in advance




calculus limits




share|cite|improve this question

























    up vote
    2
    down vote

    favorite
    2













    Calculate
    $$lim_nrightarrow infty left( arctanfrac12 + arctan frac12.2^2 +....+ arctan frac 12n^2right)$$




    My answer: i know that $$ sum_n=1^N arctan left( frac2n^2 right) =sum_n=1^N arctan (n+1)-arctan(n-1)$$



    as Im not able To find the $sum_k=1^n arctan frac 12k^2$



    I need help,,,,,any hints /solution will be aprreciated



    thanks in advance




    calculus limits




    share|cite|improve this question























      up vote
      2
      down vote

      favorite
      2









      up vote
      2
      down vote

      favorite
      2






      2






      Calculate
      $$lim_nrightarrow infty left( arctanfrac12 + arctan frac12.2^2 +....+ arctan frac 12n^2right)$$




      My answer: i know that $$ sum_n=1^N arctan left( frac2n^2 right) =sum_n=1^N arctan (n+1)-arctan(n-1)$$



      as Im not able To find the $sum_k=1^n arctan frac 12k^2$



      I need help,,,,,any hints /solution will be aprreciated



      thanks in advance




      calculus limits




      share|cite|improve this question














      Calculate
      $$lim_nrightarrow infty left( arctanfrac12 + arctan frac12.2^2 +....+ arctan frac 12n^2right)$$




      My answer: i know that $$ sum_n=1^N arctan left( frac2n^2 right) =sum_n=1^N arctan (n+1)-arctan(n-1)$$



      as Im not able To find the $sum_k=1^n arctan frac 12k^2$



      I need help,,,,,any hints /solution will be aprreciated



      thanks in advance





      calculus limits





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Jul 27 at 3:40









      Nosrati

      19.2k41544




      19.2k41544









      asked Jul 27 at 3:17









      Messi fifa

      1668




      1668




















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          8
          down vote



          accepted










          $$frac12n^2=frac24n^2=frac21+4n^2-1=frac(2n+1)-(2n-1)1+(2n+1)(2n-1)$$



          Therefore,



          $$arctan left( frac12n^2right) = arctan (2n+1) - arctan(2n-1) $$



          Can you perform the telescoping sum now?






          share|cite|improve this answer





















          • @Messififa Try to use the well known formula : $$arctan left( fracb-a1+ab right) = arctan(b) -arctan (a)$$
            – Jaideep Khare
            Jul 27 at 3:28










          • ya i understand that but after that how can i used telescope sum? can u elaborate that telescope sum
            – Messi fifa
            Jul 27 at 3:32






          • 1




            @Messififa I suggest you to write for maybe $N=4$ or $N=5$ and you will see.
            – Ovi
            Jul 27 at 3:34










          • @JaideepKhare i got $frac -pi4$ is its correct
            – Messi fifa
            Jul 27 at 4:01










          • @Messififa Well it's almost correct. Correct answer is $fracpi4$. Just the sign is opposite.
            – Jaideep Khare
            Jul 27 at 15:43











          Your Answer




          StackExchange.ifUsing("editor", function ()
          return StackExchange.using("mathjaxEditing", function ()
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          );
          );
          , "mathjax-editing");

          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',
          convertImagesToLinks: true,
          noModals: false,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          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%2f2864016%2flim-n-rightarrow-infty-arctan-frac12-arctan-frac12-22%23new-answer', 'question_page');

          );

          Post as a guest

















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          8
          down vote



          accepted










          $$frac12n^2=frac24n^2=frac21+4n^2-1=frac(2n+1)-(2n-1)1+(2n+1)(2n-1)$$



          Therefore,



          $$arctan left( frac12n^2right) = arctan (2n+1) - arctan(2n-1) $$



          Can you perform the telescoping sum now?






          share|cite|improve this answer





















          • @Messififa Try to use the well known formula : $$arctan left( fracb-a1+ab right) = arctan(b) -arctan (a)$$
            – Jaideep Khare
            Jul 27 at 3:28










          • ya i understand that but after that how can i used telescope sum? can u elaborate that telescope sum
            – Messi fifa
            Jul 27 at 3:32






          • 1




            @Messififa I suggest you to write for maybe $N=4$ or $N=5$ and you will see.
            – Ovi
            Jul 27 at 3:34










          • @JaideepKhare i got $frac -pi4$ is its correct
            – Messi fifa
            Jul 27 at 4:01










          • @Messififa Well it's almost correct. Correct answer is $fracpi4$. Just the sign is opposite.
            – Jaideep Khare
            Jul 27 at 15:43















          up vote
          8
          down vote



          accepted










          $$frac12n^2=frac24n^2=frac21+4n^2-1=frac(2n+1)-(2n-1)1+(2n+1)(2n-1)$$



          Therefore,



          $$arctan left( frac12n^2right) = arctan (2n+1) - arctan(2n-1) $$



          Can you perform the telescoping sum now?






          share|cite|improve this answer





















          • @Messififa Try to use the well known formula : $$arctan left( fracb-a1+ab right) = arctan(b) -arctan (a)$$
            – Jaideep Khare
            Jul 27 at 3:28










          • ya i understand that but after that how can i used telescope sum? can u elaborate that telescope sum
            – Messi fifa
            Jul 27 at 3:32






          • 1




            @Messififa I suggest you to write for maybe $N=4$ or $N=5$ and you will see.
            – Ovi
            Jul 27 at 3:34










          • @JaideepKhare i got $frac -pi4$ is its correct
            – Messi fifa
            Jul 27 at 4:01










          • @Messififa Well it's almost correct. Correct answer is $fracpi4$. Just the sign is opposite.
            – Jaideep Khare
            Jul 27 at 15:43













          up vote
          8
          down vote



          accepted







          up vote
          8
          down vote



          accepted






          $$frac12n^2=frac24n^2=frac21+4n^2-1=frac(2n+1)-(2n-1)1+(2n+1)(2n-1)$$



          Therefore,



          $$arctan left( frac12n^2right) = arctan (2n+1) - arctan(2n-1) $$



          Can you perform the telescoping sum now?






          share|cite|improve this answer













          $$frac12n^2=frac24n^2=frac21+4n^2-1=frac(2n+1)-(2n-1)1+(2n+1)(2n-1)$$



          Therefore,



          $$arctan left( frac12n^2right) = arctan (2n+1) - arctan(2n-1) $$



          Can you perform the telescoping sum now?







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 27 at 3:20









          Jaideep Khare

          17.6k32265




          17.6k32265











          • @Messififa Try to use the well known formula : $$arctan left( fracb-a1+ab right) = arctan(b) -arctan (a)$$
            – Jaideep Khare
            Jul 27 at 3:28










          • ya i understand that but after that how can i used telescope sum? can u elaborate that telescope sum
            – Messi fifa
            Jul 27 at 3:32






          • 1




            @Messififa I suggest you to write for maybe $N=4$ or $N=5$ and you will see.
            – Ovi
            Jul 27 at 3:34










          • @JaideepKhare i got $frac -pi4$ is its correct
            – Messi fifa
            Jul 27 at 4:01










          • @Messififa Well it's almost correct. Correct answer is $fracpi4$. Just the sign is opposite.
            – Jaideep Khare
            Jul 27 at 15:43

















          • @Messififa Try to use the well known formula : $$arctan left( fracb-a1+ab right) = arctan(b) -arctan (a)$$
            – Jaideep Khare
            Jul 27 at 3:28










          • ya i understand that but after that how can i used telescope sum? can u elaborate that telescope sum
            – Messi fifa
            Jul 27 at 3:32






          • 1




            @Messififa I suggest you to write for maybe $N=4$ or $N=5$ and you will see.
            – Ovi
            Jul 27 at 3:34










          • @JaideepKhare i got $frac -pi4$ is its correct
            – Messi fifa
            Jul 27 at 4:01










          • @Messififa Well it's almost correct. Correct answer is $fracpi4$. Just the sign is opposite.
            – Jaideep Khare
            Jul 27 at 15:43
















          @Messififa Try to use the well known formula : $$arctan left( fracb-a1+ab right) = arctan(b) -arctan (a)$$
          – Jaideep Khare
          Jul 27 at 3:28




          @Messififa Try to use the well known formula : $$arctan left( fracb-a1+ab right) = arctan(b) -arctan (a)$$
          – Jaideep Khare
          Jul 27 at 3:28












          ya i understand that but after that how can i used telescope sum? can u elaborate that telescope sum
          – Messi fifa
          Jul 27 at 3:32




          ya i understand that but after that how can i used telescope sum? can u elaborate that telescope sum
          – Messi fifa
          Jul 27 at 3:32




          1




          1




          @Messififa I suggest you to write for maybe $N=4$ or $N=5$ and you will see.
          – Ovi
          Jul 27 at 3:34




          @Messififa I suggest you to write for maybe $N=4$ or $N=5$ and you will see.
          – Ovi
          Jul 27 at 3:34












          @JaideepKhare i got $frac -pi4$ is its correct
          – Messi fifa
          Jul 27 at 4:01




          @JaideepKhare i got $frac -pi4$ is its correct
          – Messi fifa
          Jul 27 at 4:01












          @Messififa Well it's almost correct. Correct answer is $fracpi4$. Just the sign is opposite.
          – Jaideep Khare
          Jul 27 at 15:43





          @Messififa Well it's almost correct. Correct answer is $fracpi4$. Just the sign is opposite.
          – Jaideep Khare
          Jul 27 at 15:43













           

          draft saved


          draft discarded


























           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2864016%2flim-n-rightarrow-infty-arctan-frac12-arctan-frac12-22%23new-answer', 'question_page');

          );

          Post as a guest
































































          Comments

          Post a Comment

          Popular posts from this blog

          What is the equation of a 3D cone with generalised tilt?

          Image
          Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
          Read more

          Relationship between determinant of matrix and determinant of adjoint?

          Image
          Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
          Read more

          Color the edges and diagonals of a regular polygon

          Image
          Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
          Read more