A Closed form for the $sum_{n=1}^{infty}sum_{k=1}^{n}frac{1}{(25k^2+25k+4)(n-k+1)^3}$












0












$begingroup$


I'm looking for a closed form for this sequence,




$$sum_{n=1}^{infty}left(sum_{k=1}^{n}frac{1}{(25k^2+25k+4)(n-k+1)^3} right)$$




I applied convergence test. The series converges.I want to know if the series is expressed with any mathematical constant. How can we do that?






calculus sequences-and-series convergence summation closed-form







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    0












    $begingroup$


    I'm looking for a closed form for this sequence,




    $$sum_{n=1}^{infty}left(sum_{k=1}^{n}frac{1}{(25k^2+25k+4)(n-k+1)^3} right)$$




    I applied convergence test. The series converges.I want to know if the series is expressed with any mathematical constant. How can we do that?






    calculus sequences-and-series convergence summation closed-form







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      2



      $begingroup$


      I'm looking for a closed form for this sequence,




      $$sum_{n=1}^{infty}left(sum_{k=1}^{n}frac{1}{(25k^2+25k+4)(n-k+1)^3} right)$$




      I applied convergence test. The series converges.I want to know if the series is expressed with any mathematical constant. How can we do that?






      calculus sequences-and-series convergence summation closed-form







      share|cite|improve this question











      $endgroup$




      I'm looking for a closed form for this sequence,




      $$sum_{n=1}^{infty}left(sum_{k=1}^{n}frac{1}{(25k^2+25k+4)(n-k+1)^3} right)$$




      I applied convergence test. The series converges.I want to know if the series is expressed with any mathematical constant. How can we do that?






      calculus sequences-and-series convergence summation closed-form




      calculus sequences-and-series convergence summation closed-form






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 27 '18 at 20:23









      Henning Makholm

      242k17308550




      242k17308550










      asked Dec 27 '18 at 20:17









      ElementaryElementary

      360111




      360111






















          2 Answers
          2






          active

          oldest

          votes


















          7












          $begingroup$

          Proceeding in
          my usual naive way,



          $begin{array}\
          S
          &=sum_{n=1}^{infty}sum_{k=1}^{n}frac{1}{(25k^2+25k+4)(n-k+1)^3}\
          &=sum_{n=1}^{infty}sum_{k=1}^{n}frac{1}{(25k^2+25k+4)(n-k+1)^3} \
          &=sum_{k=1}^{infty}sum_{n=k}^{infty}frac{1}{(25k^2+25k+4)(n-k+1)^3} \
          &=sum_{k=1}^{infty}frac1{(25k^2+25k+4)}sum_{n=k}^{infty}frac{1}{(n-k+1)^3} \
          &=sum_{k=1}^{infty}frac1{(25k^2+25k+4)}sum_{n=1}^{infty}frac{1}{n^3} \
          &=zeta(3)sum_{k=1}^{infty}frac1{(25k^2+25k+4)}\
          &=zeta(3)sum_{k=1}^{infty}frac1{(5k+1)(5k+4)}\
          &=zeta(3)sum_{k=1}^{infty}frac13left(frac1{5k+1}-frac1{5k+4}right)\
          &=frac{zeta(3)}{3}lim_{m to infty} sum_{k=1}^{m}left(frac1{5k+1}-frac1{5k+4}right)\
          &=frac{zeta(3)}{15}lim_{m to infty} left(sum_{k=1}^{m}frac1{k+1/5}-sum_{k=1}^{m}frac1{k+4/5}right)\
          &=-frac{zeta(3)}{15}lim_{m to infty} left(sum_{k=1}^{m}(frac1{k}-frac1{k+1/5})-sum_{k=1}^{m}(frac1{k}-frac1{k+4/5})right)\
          &=-frac{zeta(3)}{15}(psi(6/5)-psi(9/5))\
          &=frac{zeta(3)}{15}(psi(9/5)-psi(6/5))\
          end{array}
          $



          where
          $psi(x)$
          is the digamma function
          (https://en.wikipedia.org/wiki/Digamma_function).



          Note:
          Wolfy says that
          $sum_{k=1}^{infty}frac1{(5k+1)(5k+4)}
          = frac{pi}{15}sqrt{1 + frac{2}{sqrt{5}}} - frac1{4}
          $          .






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            (+1) Thank you for answer..
            $endgroup$
            – Elementary
            Dec 27 '18 at 21:01






          • 2




            $begingroup$
            First, there is a typo $frac{1}{(5k+1)(5k+4)} = color{red}{frac13} left(frac{1}{5k+1} - frac{1}{5k+4}right)$. Second, you can use identities $$psi(z+1) = psi(z) + frac1zquadtext{ and }quadpsi(1-z) - psi(z) = picotpi z$$ to get rid of $psi$.
            $endgroup$
            – achille hui
            Dec 27 '18 at 21:04






          • 1




            $begingroup$
            Can we do what Wolfy did?
            $endgroup$
            – Elementary
            Dec 27 '18 at 21:04






          • 1




            $begingroup$
            @achille hui: Thanks for the catch and the reference to the digamma identity.
            $endgroup$
            – marty cohen
            Dec 27 '18 at 22:50



















          4












          $begingroup$

          Change the order of summation, so it's $sum_{k=1}^infty sum_{n=k}^infty$.
          Then I get
          $$ {frac {zeta left( 3 right) left( 4,pi,cot left( pi/5
          right) -15 right) }{60}}
          $$
          You could also write $$cot(pi/5) = frac{sqrt{2}}{20} (5 + sqrt{5})^{3/2}$$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            I just stopped at the digamma function. Didn't realize it could be turned into a trig function.
            $endgroup$
            – marty cohen
            Dec 27 '18 at 20:47










          • $begingroup$
            $cot(pi/5)=sqrt{1+2/sqrt{5}}$ is a little bit simpler.
            $endgroup$
            – metamorphy
            Dec 27 '18 at 20:56










          • $begingroup$
            That's what Wolfy says too.
            $endgroup$
            – marty cohen
            Dec 27 '18 at 20:56










          • $begingroup$
            Dear teacher, is it possible to write which method you used? How do you get the Closed Form..? Thank You.
            $endgroup$
            – Elementary
            Dec 27 '18 at 21:00













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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          7












          $begingroup$

          Proceeding in
          my usual naive way,



          $begin{array}\
          S
          &=sum_{n=1}^{infty}sum_{k=1}^{n}frac{1}{(25k^2+25k+4)(n-k+1)^3}\
          &=sum_{n=1}^{infty}sum_{k=1}^{n}frac{1}{(25k^2+25k+4)(n-k+1)^3} \
          &=sum_{k=1}^{infty}sum_{n=k}^{infty}frac{1}{(25k^2+25k+4)(n-k+1)^3} \
          &=sum_{k=1}^{infty}frac1{(25k^2+25k+4)}sum_{n=k}^{infty}frac{1}{(n-k+1)^3} \
          &=sum_{k=1}^{infty}frac1{(25k^2+25k+4)}sum_{n=1}^{infty}frac{1}{n^3} \
          &=zeta(3)sum_{k=1}^{infty}frac1{(25k^2+25k+4)}\
          &=zeta(3)sum_{k=1}^{infty}frac1{(5k+1)(5k+4)}\
          &=zeta(3)sum_{k=1}^{infty}frac13left(frac1{5k+1}-frac1{5k+4}right)\
          &=frac{zeta(3)}{3}lim_{m to infty} sum_{k=1}^{m}left(frac1{5k+1}-frac1{5k+4}right)\
          &=frac{zeta(3)}{15}lim_{m to infty} left(sum_{k=1}^{m}frac1{k+1/5}-sum_{k=1}^{m}frac1{k+4/5}right)\
          &=-frac{zeta(3)}{15}lim_{m to infty} left(sum_{k=1}^{m}(frac1{k}-frac1{k+1/5})-sum_{k=1}^{m}(frac1{k}-frac1{k+4/5})right)\
          &=-frac{zeta(3)}{15}(psi(6/5)-psi(9/5))\
          &=frac{zeta(3)}{15}(psi(9/5)-psi(6/5))\
          end{array}
          $



          where
          $psi(x)$
          is the digamma function
          (https://en.wikipedia.org/wiki/Digamma_function).



          Note:
          Wolfy says that
          $sum_{k=1}^{infty}frac1{(5k+1)(5k+4)}
          = frac{pi}{15}sqrt{1 + frac{2}{sqrt{5}}} - frac1{4}
          $          .






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            (+1) Thank you for answer..
            $endgroup$
            – Elementary
            Dec 27 '18 at 21:01






          • 2




            $begingroup$
            First, there is a typo $frac{1}{(5k+1)(5k+4)} = color{red}{frac13} left(frac{1}{5k+1} - frac{1}{5k+4}right)$. Second, you can use identities $$psi(z+1) = psi(z) + frac1zquadtext{ and }quadpsi(1-z) - psi(z) = picotpi z$$ to get rid of $psi$.
            $endgroup$
            – achille hui
            Dec 27 '18 at 21:04






          • 1




            $begingroup$
            Can we do what Wolfy did?
            $endgroup$
            – Elementary
            Dec 27 '18 at 21:04






          • 1




            $begingroup$
            @achille hui: Thanks for the catch and the reference to the digamma identity.
            $endgroup$
            – marty cohen
            Dec 27 '18 at 22:50
















          7












          $begingroup$

          Proceeding in
          my usual naive way,



          $begin{array}\
          S
          &=sum_{n=1}^{infty}sum_{k=1}^{n}frac{1}{(25k^2+25k+4)(n-k+1)^3}\
          &=sum_{n=1}^{infty}sum_{k=1}^{n}frac{1}{(25k^2+25k+4)(n-k+1)^3} \
          &=sum_{k=1}^{infty}sum_{n=k}^{infty}frac{1}{(25k^2+25k+4)(n-k+1)^3} \
          &=sum_{k=1}^{infty}frac1{(25k^2+25k+4)}sum_{n=k}^{infty}frac{1}{(n-k+1)^3} \
          &=sum_{k=1}^{infty}frac1{(25k^2+25k+4)}sum_{n=1}^{infty}frac{1}{n^3} \
          &=zeta(3)sum_{k=1}^{infty}frac1{(25k^2+25k+4)}\
          &=zeta(3)sum_{k=1}^{infty}frac1{(5k+1)(5k+4)}\
          &=zeta(3)sum_{k=1}^{infty}frac13left(frac1{5k+1}-frac1{5k+4}right)\
          &=frac{zeta(3)}{3}lim_{m to infty} sum_{k=1}^{m}left(frac1{5k+1}-frac1{5k+4}right)\
          &=frac{zeta(3)}{15}lim_{m to infty} left(sum_{k=1}^{m}frac1{k+1/5}-sum_{k=1}^{m}frac1{k+4/5}right)\
          &=-frac{zeta(3)}{15}lim_{m to infty} left(sum_{k=1}^{m}(frac1{k}-frac1{k+1/5})-sum_{k=1}^{m}(frac1{k}-frac1{k+4/5})right)\
          &=-frac{zeta(3)}{15}(psi(6/5)-psi(9/5))\
          &=frac{zeta(3)}{15}(psi(9/5)-psi(6/5))\
          end{array}
          $



          where
          $psi(x)$
          is the digamma function
          (https://en.wikipedia.org/wiki/Digamma_function).



          Note:
          Wolfy says that
          $sum_{k=1}^{infty}frac1{(5k+1)(5k+4)}
          = frac{pi}{15}sqrt{1 + frac{2}{sqrt{5}}} - frac1{4}
          $          .






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            (+1) Thank you for answer..
            $endgroup$
            – Elementary
            Dec 27 '18 at 21:01






          • 2




            $begingroup$
            First, there is a typo $frac{1}{(5k+1)(5k+4)} = color{red}{frac13} left(frac{1}{5k+1} - frac{1}{5k+4}right)$. Second, you can use identities $$psi(z+1) = psi(z) + frac1zquadtext{ and }quadpsi(1-z) - psi(z) = picotpi z$$ to get rid of $psi$.
            $endgroup$
            – achille hui
            Dec 27 '18 at 21:04






          • 1




            $begingroup$
            Can we do what Wolfy did?
            $endgroup$
            – Elementary
            Dec 27 '18 at 21:04






          • 1




            $begingroup$
            @achille hui: Thanks for the catch and the reference to the digamma identity.
            $endgroup$
            – marty cohen
            Dec 27 '18 at 22:50














          7












          7








          7





          $begingroup$

          Proceeding in
          my usual naive way,



          $begin{array}\
          S
          &=sum_{n=1}^{infty}sum_{k=1}^{n}frac{1}{(25k^2+25k+4)(n-k+1)^3}\
          &=sum_{n=1}^{infty}sum_{k=1}^{n}frac{1}{(25k^2+25k+4)(n-k+1)^3} \
          &=sum_{k=1}^{infty}sum_{n=k}^{infty}frac{1}{(25k^2+25k+4)(n-k+1)^3} \
          &=sum_{k=1}^{infty}frac1{(25k^2+25k+4)}sum_{n=k}^{infty}frac{1}{(n-k+1)^3} \
          &=sum_{k=1}^{infty}frac1{(25k^2+25k+4)}sum_{n=1}^{infty}frac{1}{n^3} \
          &=zeta(3)sum_{k=1}^{infty}frac1{(25k^2+25k+4)}\
          &=zeta(3)sum_{k=1}^{infty}frac1{(5k+1)(5k+4)}\
          &=zeta(3)sum_{k=1}^{infty}frac13left(frac1{5k+1}-frac1{5k+4}right)\
          &=frac{zeta(3)}{3}lim_{m to infty} sum_{k=1}^{m}left(frac1{5k+1}-frac1{5k+4}right)\
          &=frac{zeta(3)}{15}lim_{m to infty} left(sum_{k=1}^{m}frac1{k+1/5}-sum_{k=1}^{m}frac1{k+4/5}right)\
          &=-frac{zeta(3)}{15}lim_{m to infty} left(sum_{k=1}^{m}(frac1{k}-frac1{k+1/5})-sum_{k=1}^{m}(frac1{k}-frac1{k+4/5})right)\
          &=-frac{zeta(3)}{15}(psi(6/5)-psi(9/5))\
          &=frac{zeta(3)}{15}(psi(9/5)-psi(6/5))\
          end{array}
          $



          where
          $psi(x)$
          is the digamma function
          (https://en.wikipedia.org/wiki/Digamma_function).



          Note:
          Wolfy says that
          $sum_{k=1}^{infty}frac1{(5k+1)(5k+4)}
          = frac{pi}{15}sqrt{1 + frac{2}{sqrt{5}}} - frac1{4}
          $          .






          share|cite|improve this answer











          $endgroup$



          Proceeding in
          my usual naive way,



          $begin{array}\
          S
          &=sum_{n=1}^{infty}sum_{k=1}^{n}frac{1}{(25k^2+25k+4)(n-k+1)^3}\
          &=sum_{n=1}^{infty}sum_{k=1}^{n}frac{1}{(25k^2+25k+4)(n-k+1)^3} \
          &=sum_{k=1}^{infty}sum_{n=k}^{infty}frac{1}{(25k^2+25k+4)(n-k+1)^3} \
          &=sum_{k=1}^{infty}frac1{(25k^2+25k+4)}sum_{n=k}^{infty}frac{1}{(n-k+1)^3} \
          &=sum_{k=1}^{infty}frac1{(25k^2+25k+4)}sum_{n=1}^{infty}frac{1}{n^3} \
          &=zeta(3)sum_{k=1}^{infty}frac1{(25k^2+25k+4)}\
          &=zeta(3)sum_{k=1}^{infty}frac1{(5k+1)(5k+4)}\
          &=zeta(3)sum_{k=1}^{infty}frac13left(frac1{5k+1}-frac1{5k+4}right)\
          &=frac{zeta(3)}{3}lim_{m to infty} sum_{k=1}^{m}left(frac1{5k+1}-frac1{5k+4}right)\
          &=frac{zeta(3)}{15}lim_{m to infty} left(sum_{k=1}^{m}frac1{k+1/5}-sum_{k=1}^{m}frac1{k+4/5}right)\
          &=-frac{zeta(3)}{15}lim_{m to infty} left(sum_{k=1}^{m}(frac1{k}-frac1{k+1/5})-sum_{k=1}^{m}(frac1{k}-frac1{k+4/5})right)\
          &=-frac{zeta(3)}{15}(psi(6/5)-psi(9/5))\
          &=frac{zeta(3)}{15}(psi(9/5)-psi(6/5))\
          end{array}
          $



          where
          $psi(x)$
          is the digamma function
          (https://en.wikipedia.org/wiki/Digamma_function).



          Note:
          Wolfy says that
          $sum_{k=1}^{infty}frac1{(5k+1)(5k+4)}
          = frac{pi}{15}sqrt{1 + frac{2}{sqrt{5}}} - frac1{4}
          $          .







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 27 '18 at 22:49

























          answered Dec 27 '18 at 20:46









          marty cohenmarty cohen

          74.4k549129




          74.4k549129








          • 1




            $begingroup$
            (+1) Thank you for answer..
            $endgroup$
            – Elementary
            Dec 27 '18 at 21:01






          • 2




            $begingroup$
            First, there is a typo $frac{1}{(5k+1)(5k+4)} = color{red}{frac13} left(frac{1}{5k+1} - frac{1}{5k+4}right)$. Second, you can use identities $$psi(z+1) = psi(z) + frac1zquadtext{ and }quadpsi(1-z) - psi(z) = picotpi z$$ to get rid of $psi$.
            $endgroup$
            – achille hui
            Dec 27 '18 at 21:04






          • 1




            $begingroup$
            Can we do what Wolfy did?
            $endgroup$
            – Elementary
            Dec 27 '18 at 21:04






          • 1




            $begingroup$
            @achille hui: Thanks for the catch and the reference to the digamma identity.
            $endgroup$
            – marty cohen
            Dec 27 '18 at 22:50














          • 1




            $begingroup$
            (+1) Thank you for answer..
            $endgroup$
            – Elementary
            Dec 27 '18 at 21:01






          • 2




            $begingroup$
            First, there is a typo $frac{1}{(5k+1)(5k+4)} = color{red}{frac13} left(frac{1}{5k+1} - frac{1}{5k+4}right)$. Second, you can use identities $$psi(z+1) = psi(z) + frac1zquadtext{ and }quadpsi(1-z) - psi(z) = picotpi z$$ to get rid of $psi$.
            $endgroup$
            – achille hui
            Dec 27 '18 at 21:04






          • 1




            $begingroup$
            Can we do what Wolfy did?
            $endgroup$
            – Elementary
            Dec 27 '18 at 21:04






          • 1




            $begingroup$
            @achille hui: Thanks for the catch and the reference to the digamma identity.
            $endgroup$
            – marty cohen
            Dec 27 '18 at 22:50








          1




          1




          $begingroup$
          (+1) Thank you for answer..
          $endgroup$
          – Elementary
          Dec 27 '18 at 21:01




          $begingroup$
          (+1) Thank you for answer..
          $endgroup$
          – Elementary
          Dec 27 '18 at 21:01




          2




          2




          $begingroup$
          First, there is a typo $frac{1}{(5k+1)(5k+4)} = color{red}{frac13} left(frac{1}{5k+1} - frac{1}{5k+4}right)$. Second, you can use identities $$psi(z+1) = psi(z) + frac1zquadtext{ and }quadpsi(1-z) - psi(z) = picotpi z$$ to get rid of $psi$.
          $endgroup$
          – achille hui
          Dec 27 '18 at 21:04




          $begingroup$
          First, there is a typo $frac{1}{(5k+1)(5k+4)} = color{red}{frac13} left(frac{1}{5k+1} - frac{1}{5k+4}right)$. Second, you can use identities $$psi(z+1) = psi(z) + frac1zquadtext{ and }quadpsi(1-z) - psi(z) = picotpi z$$ to get rid of $psi$.
          $endgroup$
          – achille hui
          Dec 27 '18 at 21:04




          1




          1




          $begingroup$
          Can we do what Wolfy did?
          $endgroup$
          – Elementary
          Dec 27 '18 at 21:04




          $begingroup$
          Can we do what Wolfy did?
          $endgroup$
          – Elementary
          Dec 27 '18 at 21:04




          1




          1




          $begingroup$
          @achille hui: Thanks for the catch and the reference to the digamma identity.
          $endgroup$
          – marty cohen
          Dec 27 '18 at 22:50




          $begingroup$
          @achille hui: Thanks for the catch and the reference to the digamma identity.
          $endgroup$
          – marty cohen
          Dec 27 '18 at 22:50











          4












          $begingroup$

          Change the order of summation, so it's $sum_{k=1}^infty sum_{n=k}^infty$.
          Then I get
          $$ {frac {zeta left( 3 right) left( 4,pi,cot left( pi/5
          right) -15 right) }{60}}
          $$
          You could also write $$cot(pi/5) = frac{sqrt{2}}{20} (5 + sqrt{5})^{3/2}$$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            I just stopped at the digamma function. Didn't realize it could be turned into a trig function.
            $endgroup$
            – marty cohen
            Dec 27 '18 at 20:47










          • $begingroup$
            $cot(pi/5)=sqrt{1+2/sqrt{5}}$ is a little bit simpler.
            $endgroup$
            – metamorphy
            Dec 27 '18 at 20:56










          • $begingroup$
            That's what Wolfy says too.
            $endgroup$
            – marty cohen
            Dec 27 '18 at 20:56










          • $begingroup$
            Dear teacher, is it possible to write which method you used? How do you get the Closed Form..? Thank You.
            $endgroup$
            – Elementary
            Dec 27 '18 at 21:00


















          4












          $begingroup$

          Change the order of summation, so it's $sum_{k=1}^infty sum_{n=k}^infty$.
          Then I get
          $$ {frac {zeta left( 3 right) left( 4,pi,cot left( pi/5
          right) -15 right) }{60}}
          $$
          You could also write $$cot(pi/5) = frac{sqrt{2}}{20} (5 + sqrt{5})^{3/2}$$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            I just stopped at the digamma function. Didn't realize it could be turned into a trig function.
            $endgroup$
            – marty cohen
            Dec 27 '18 at 20:47










          • $begingroup$
            $cot(pi/5)=sqrt{1+2/sqrt{5}}$ is a little bit simpler.
            $endgroup$
            – metamorphy
            Dec 27 '18 at 20:56










          • $begingroup$
            That's what Wolfy says too.
            $endgroup$
            – marty cohen
            Dec 27 '18 at 20:56










          • $begingroup$
            Dear teacher, is it possible to write which method you used? How do you get the Closed Form..? Thank You.
            $endgroup$
            – Elementary
            Dec 27 '18 at 21:00
















          4












          4








          4





          $begingroup$

          Change the order of summation, so it's $sum_{k=1}^infty sum_{n=k}^infty$.
          Then I get
          $$ {frac {zeta left( 3 right) left( 4,pi,cot left( pi/5
          right) -15 right) }{60}}
          $$
          You could also write $$cot(pi/5) = frac{sqrt{2}}{20} (5 + sqrt{5})^{3/2}$$






          share|cite|improve this answer









          $endgroup$



          Change the order of summation, so it's $sum_{k=1}^infty sum_{n=k}^infty$.
          Then I get
          $$ {frac {zeta left( 3 right) left( 4,pi,cot left( pi/5
          right) -15 right) }{60}}
          $$
          You could also write $$cot(pi/5) = frac{sqrt{2}}{20} (5 + sqrt{5})^{3/2}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 27 '18 at 20:43









          Robert IsraelRobert Israel

          327k23216470




          327k23216470








          • 1




            $begingroup$
            I just stopped at the digamma function. Didn't realize it could be turned into a trig function.
            $endgroup$
            – marty cohen
            Dec 27 '18 at 20:47










          • $begingroup$
            $cot(pi/5)=sqrt{1+2/sqrt{5}}$ is a little bit simpler.
            $endgroup$
            – metamorphy
            Dec 27 '18 at 20:56










          • $begingroup$
            That's what Wolfy says too.
            $endgroup$
            – marty cohen
            Dec 27 '18 at 20:56










          • $begingroup$
            Dear teacher, is it possible to write which method you used? How do you get the Closed Form..? Thank You.
            $endgroup$
            – Elementary
            Dec 27 '18 at 21:00
















          • 1




            $begingroup$
            I just stopped at the digamma function. Didn't realize it could be turned into a trig function.
            $endgroup$
            – marty cohen
            Dec 27 '18 at 20:47










          • $begingroup$
            $cot(pi/5)=sqrt{1+2/sqrt{5}}$ is a little bit simpler.
            $endgroup$
            – metamorphy
            Dec 27 '18 at 20:56










          • $begingroup$
            That's what Wolfy says too.
            $endgroup$
            – marty cohen
            Dec 27 '18 at 20:56










          • $begingroup$
            Dear teacher, is it possible to write which method you used? How do you get the Closed Form..? Thank You.
            $endgroup$
            – Elementary
            Dec 27 '18 at 21:00










          1




          1




          $begingroup$
          I just stopped at the digamma function. Didn't realize it could be turned into a trig function.
          $endgroup$
          – marty cohen
          Dec 27 '18 at 20:47




          $begingroup$
          I just stopped at the digamma function. Didn't realize it could be turned into a trig function.
          $endgroup$
          – marty cohen
          Dec 27 '18 at 20:47












          $begingroup$
          $cot(pi/5)=sqrt{1+2/sqrt{5}}$ is a little bit simpler.
          $endgroup$
          – metamorphy
          Dec 27 '18 at 20:56




          $begingroup$
          $cot(pi/5)=sqrt{1+2/sqrt{5}}$ is a little bit simpler.
          $endgroup$
          – metamorphy
          Dec 27 '18 at 20:56












          $begingroup$
          That's what Wolfy says too.
          $endgroup$
          – marty cohen
          Dec 27 '18 at 20:56




          $begingroup$
          That's what Wolfy says too.
          $endgroup$
          – marty cohen
          Dec 27 '18 at 20:56












          $begingroup$
          Dear teacher, is it possible to write which method you used? How do you get the Closed Form..? Thank You.
          $endgroup$
          – Elementary
          Dec 27 '18 at 21:00






          $begingroup$
          Dear teacher, is it possible to write which method you used? How do you get the Closed Form..? Thank You.
          $endgroup$
          – Elementary
          Dec 27 '18 at 21:00




















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