Proof that $ sum_{n=2}^{infty} frac{2}{3^n cdot (n^3-n)} = frac{-1}{2} + frac{4}{3}sum_{n=1}^{infty}...












1












$begingroup$


Task



Proof that $ sum_{n=2}^{infty} frac{2}{3^n cdot (n^3-n)} = -frac{1}{2} + frac{4}{3}sum_{n=1}^{infty} frac{1}{n cdot 3^n}$



About



Hi, I have been trying to solve this task since yesterday. I have idea that I can evaluate both sides and show that they are the same. I supposed $ sum_{n=1}^{infty} frac{1}{n cdot 3^n} $ there is formula for that.
So I want to evaluate in the same way left side. So I compute that



$$ sum_{n=2}^{infty} frac{2}{3^n cdot (n^3-n)} = sum_{n=2}^{infty} frac{1}{3^n cdot n (n-1)} - frac{1}{3^n cdot n (n+1)} $$
I am trying to transform it to use that formula:
$$ sum_{n=1}^{infty} frac{1}{ncdot p^n} = lnfrac{p}{p-1} $$
but I still defeat.
So please tell me, there are better ways to proof that or should I consider to change my field of study?






real-analysis proof-writing







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Task



    Proof that $ sum_{n=2}^{infty} frac{2}{3^n cdot (n^3-n)} = -frac{1}{2} + frac{4}{3}sum_{n=1}^{infty} frac{1}{n cdot 3^n}$



    About



    Hi, I have been trying to solve this task since yesterday. I have idea that I can evaluate both sides and show that they are the same. I supposed $ sum_{n=1}^{infty} frac{1}{n cdot 3^n} $ there is formula for that.
    So I want to evaluate in the same way left side. So I compute that



    $$ sum_{n=2}^{infty} frac{2}{3^n cdot (n^3-n)} = sum_{n=2}^{infty} frac{1}{3^n cdot n (n-1)} - frac{1}{3^n cdot n (n+1)} $$
    I am trying to transform it to use that formula:
    $$ sum_{n=1}^{infty} frac{1}{ncdot p^n} = lnfrac{p}{p-1} $$
    but I still defeat.
    So please tell me, there are better ways to proof that or should I consider to change my field of study?






    real-analysis proof-writing







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Task



      Proof that $ sum_{n=2}^{infty} frac{2}{3^n cdot (n^3-n)} = -frac{1}{2} + frac{4}{3}sum_{n=1}^{infty} frac{1}{n cdot 3^n}$



      About



      Hi, I have been trying to solve this task since yesterday. I have idea that I can evaluate both sides and show that they are the same. I supposed $ sum_{n=1}^{infty} frac{1}{n cdot 3^n} $ there is formula for that.
      So I want to evaluate in the same way left side. So I compute that



      $$ sum_{n=2}^{infty} frac{2}{3^n cdot (n^3-n)} = sum_{n=2}^{infty} frac{1}{3^n cdot n (n-1)} - frac{1}{3^n cdot n (n+1)} $$
      I am trying to transform it to use that formula:
      $$ sum_{n=1}^{infty} frac{1}{ncdot p^n} = lnfrac{p}{p-1} $$
      but I still defeat.
      So please tell me, there are better ways to proof that or should I consider to change my field of study?






      real-analysis proof-writing







      share|cite|improve this question









      $endgroup$




      Task



      Proof that $ sum_{n=2}^{infty} frac{2}{3^n cdot (n^3-n)} = -frac{1}{2} + frac{4}{3}sum_{n=1}^{infty} frac{1}{n cdot 3^n}$



      About



      Hi, I have been trying to solve this task since yesterday. I have idea that I can evaluate both sides and show that they are the same. I supposed $ sum_{n=1}^{infty} frac{1}{n cdot 3^n} $ there is formula for that.
      So I want to evaluate in the same way left side. So I compute that



      $$ sum_{n=2}^{infty} frac{2}{3^n cdot (n^3-n)} = sum_{n=2}^{infty} frac{1}{3^n cdot n (n-1)} - frac{1}{3^n cdot n (n+1)} $$
      I am trying to transform it to use that formula:
      $$ sum_{n=1}^{infty} frac{1}{ncdot p^n} = lnfrac{p}{p-1} $$
      but I still defeat.
      So please tell me, there are better ways to proof that or should I consider to change my field of study?






      real-analysis proof-writing




      real-analysis proof-writing






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 19 '18 at 19:29









      VirtualUserVirtualUser

      1,096117




      1,096117






















          2 Answers
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          1












          $begingroup$

          Hints:



          $$n^3 - n = (n-1) n (n+1)$$



          $$frac{1}{n (n+1)} = frac{1}{n} - frac{1}{n+1}$$
          $$frac{1}{(n-1) n} = frac{1}{n-1} - frac{1}{n}$$
          $$frac{1}{(n-1)(n+1)} = frac{1/2}{n-1} - frac{1/2}{n+1}$$
          $$frac{1}{(n-1)n(n+1)} = frac{1}{(n-1)n} - frac{1}{(n-1)(n+1)} = frac{1}{n-1} - frac{1}{n} - frac{1/2}{n-1} + frac{1/2}{n+1} = frac{1/2}{n-1} - frac{1}{n} + frac{1/2}{n+1}$$



          $$frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n}
          + frac{1}{3^n (n+1)}$$



          Then try to combine terms when substituting this into the sum.




          There is probably a more elegant way to do the following computations... $$begin{align}sum_{n ge 2} frac{2}{3^n (n-1) n (n+1)} &= frac{1}{3^2} + left(frac{1}{3^3} - frac{2}{3^2}right)frac{1}{2} + sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} \ &= frac{1}{54} + left(3 - 2 + frac{1}{3}right)sum_{n ge 3} frac{1}{3^n n} \ &= frac{1}{54} + frac{4}{3} sum_{nge 3} frac{1}{3^n n} \ &= frac{1}{54} - frac{4}{3} left(frac{1}{3} + frac{1}{9 cdot 2} right) + frac{4}{3} sum_{nge 1} frac{1}{3^n n} \ &= - frac{1}{2} + frac{4}{3} sum_{nge 1} frac{1}{3^n n}. end{align}$$







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you explain how you get this? $ sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} $? I know that $ frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)} $ but you also take $1/n$ before and it transformed into factors like that: $ frac{1}{3^{n-1}} $
            $endgroup$
            – VirtualUser
            Dec 19 '18 at 20:36












          • $begingroup$
            @VirtualUser Try writing out the first few terms of the sum $sum_{n ge 2} left(frac{1}{3^n(n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)}right)$ explicitly and it should become clear.
            $endgroup$
            – angryavian
            Dec 20 '18 at 16:45










          • $begingroup$
            You have just sorted elements by n'th term, right?
            $endgroup$
            – VirtualUser
            Dec 20 '18 at 18:20



















          2












          $begingroup$

          hint



          $$frac{2}{n^3-n}=frac{-2}{n}+frac{1}{n-1}+frac{1}{n+1}$$






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            2 Answers
            2






            active

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            active

            oldest

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            active

            oldest

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            1












            $begingroup$

            Hints:



            $$n^3 - n = (n-1) n (n+1)$$



            $$frac{1}{n (n+1)} = frac{1}{n} - frac{1}{n+1}$$
            $$frac{1}{(n-1) n} = frac{1}{n-1} - frac{1}{n}$$
            $$frac{1}{(n-1)(n+1)} = frac{1/2}{n-1} - frac{1/2}{n+1}$$
            $$frac{1}{(n-1)n(n+1)} = frac{1}{(n-1)n} - frac{1}{(n-1)(n+1)} = frac{1}{n-1} - frac{1}{n} - frac{1/2}{n-1} + frac{1/2}{n+1} = frac{1/2}{n-1} - frac{1}{n} + frac{1/2}{n+1}$$



            $$frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n}
            + frac{1}{3^n (n+1)}$$



            Then try to combine terms when substituting this into the sum.




            There is probably a more elegant way to do the following computations... $$begin{align}sum_{n ge 2} frac{2}{3^n (n-1) n (n+1)} &= frac{1}{3^2} + left(frac{1}{3^3} - frac{2}{3^2}right)frac{1}{2} + sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} \ &= frac{1}{54} + left(3 - 2 + frac{1}{3}right)sum_{n ge 3} frac{1}{3^n n} \ &= frac{1}{54} + frac{4}{3} sum_{nge 3} frac{1}{3^n n} \ &= frac{1}{54} - frac{4}{3} left(frac{1}{3} + frac{1}{9 cdot 2} right) + frac{4}{3} sum_{nge 1} frac{1}{3^n n} \ &= - frac{1}{2} + frac{4}{3} sum_{nge 1} frac{1}{3^n n}. end{align}$$







            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Can you explain how you get this? $ sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} $? I know that $ frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)} $ but you also take $1/n$ before and it transformed into factors like that: $ frac{1}{3^{n-1}} $
              $endgroup$
              – VirtualUser
              Dec 19 '18 at 20:36












            • $begingroup$
              @VirtualUser Try writing out the first few terms of the sum $sum_{n ge 2} left(frac{1}{3^n(n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)}right)$ explicitly and it should become clear.
              $endgroup$
              – angryavian
              Dec 20 '18 at 16:45










            • $begingroup$
              You have just sorted elements by n'th term, right?
              $endgroup$
              – VirtualUser
              Dec 20 '18 at 18:20
















            1












            $begingroup$

            Hints:



            $$n^3 - n = (n-1) n (n+1)$$



            $$frac{1}{n (n+1)} = frac{1}{n} - frac{1}{n+1}$$
            $$frac{1}{(n-1) n} = frac{1}{n-1} - frac{1}{n}$$
            $$frac{1}{(n-1)(n+1)} = frac{1/2}{n-1} - frac{1/2}{n+1}$$
            $$frac{1}{(n-1)n(n+1)} = frac{1}{(n-1)n} - frac{1}{(n-1)(n+1)} = frac{1}{n-1} - frac{1}{n} - frac{1/2}{n-1} + frac{1/2}{n+1} = frac{1/2}{n-1} - frac{1}{n} + frac{1/2}{n+1}$$



            $$frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n}
            + frac{1}{3^n (n+1)}$$



            Then try to combine terms when substituting this into the sum.




            There is probably a more elegant way to do the following computations... $$begin{align}sum_{n ge 2} frac{2}{3^n (n-1) n (n+1)} &= frac{1}{3^2} + left(frac{1}{3^3} - frac{2}{3^2}right)frac{1}{2} + sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} \ &= frac{1}{54} + left(3 - 2 + frac{1}{3}right)sum_{n ge 3} frac{1}{3^n n} \ &= frac{1}{54} + frac{4}{3} sum_{nge 3} frac{1}{3^n n} \ &= frac{1}{54} - frac{4}{3} left(frac{1}{3} + frac{1}{9 cdot 2} right) + frac{4}{3} sum_{nge 1} frac{1}{3^n n} \ &= - frac{1}{2} + frac{4}{3} sum_{nge 1} frac{1}{3^n n}. end{align}$$







            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Can you explain how you get this? $ sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} $? I know that $ frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)} $ but you also take $1/n$ before and it transformed into factors like that: $ frac{1}{3^{n-1}} $
              $endgroup$
              – VirtualUser
              Dec 19 '18 at 20:36












            • $begingroup$
              @VirtualUser Try writing out the first few terms of the sum $sum_{n ge 2} left(frac{1}{3^n(n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)}right)$ explicitly and it should become clear.
              $endgroup$
              – angryavian
              Dec 20 '18 at 16:45










            • $begingroup$
              You have just sorted elements by n'th term, right?
              $endgroup$
              – VirtualUser
              Dec 20 '18 at 18:20














            1












            1








            1





            $begingroup$

            Hints:



            $$n^3 - n = (n-1) n (n+1)$$



            $$frac{1}{n (n+1)} = frac{1}{n} - frac{1}{n+1}$$
            $$frac{1}{(n-1) n} = frac{1}{n-1} - frac{1}{n}$$
            $$frac{1}{(n-1)(n+1)} = frac{1/2}{n-1} - frac{1/2}{n+1}$$
            $$frac{1}{(n-1)n(n+1)} = frac{1}{(n-1)n} - frac{1}{(n-1)(n+1)} = frac{1}{n-1} - frac{1}{n} - frac{1/2}{n-1} + frac{1/2}{n+1} = frac{1/2}{n-1} - frac{1}{n} + frac{1/2}{n+1}$$



            $$frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n}
            + frac{1}{3^n (n+1)}$$



            Then try to combine terms when substituting this into the sum.




            There is probably a more elegant way to do the following computations... $$begin{align}sum_{n ge 2} frac{2}{3^n (n-1) n (n+1)} &= frac{1}{3^2} + left(frac{1}{3^3} - frac{2}{3^2}right)frac{1}{2} + sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} \ &= frac{1}{54} + left(3 - 2 + frac{1}{3}right)sum_{n ge 3} frac{1}{3^n n} \ &= frac{1}{54} + frac{4}{3} sum_{nge 3} frac{1}{3^n n} \ &= frac{1}{54} - frac{4}{3} left(frac{1}{3} + frac{1}{9 cdot 2} right) + frac{4}{3} sum_{nge 1} frac{1}{3^n n} \ &= - frac{1}{2} + frac{4}{3} sum_{nge 1} frac{1}{3^n n}. end{align}$$







            share|cite|improve this answer











            $endgroup$



            Hints:



            $$n^3 - n = (n-1) n (n+1)$$



            $$frac{1}{n (n+1)} = frac{1}{n} - frac{1}{n+1}$$
            $$frac{1}{(n-1) n} = frac{1}{n-1} - frac{1}{n}$$
            $$frac{1}{(n-1)(n+1)} = frac{1/2}{n-1} - frac{1/2}{n+1}$$
            $$frac{1}{(n-1)n(n+1)} = frac{1}{(n-1)n} - frac{1}{(n-1)(n+1)} = frac{1}{n-1} - frac{1}{n} - frac{1/2}{n-1} + frac{1/2}{n+1} = frac{1/2}{n-1} - frac{1}{n} + frac{1/2}{n+1}$$



            $$frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n}
            + frac{1}{3^n (n+1)}$$



            Then try to combine terms when substituting this into the sum.




            There is probably a more elegant way to do the following computations... $$begin{align}sum_{n ge 2} frac{2}{3^n (n-1) n (n+1)} &= frac{1}{3^2} + left(frac{1}{3^3} - frac{2}{3^2}right)frac{1}{2} + sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} \ &= frac{1}{54} + left(3 - 2 + frac{1}{3}right)sum_{n ge 3} frac{1}{3^n n} \ &= frac{1}{54} + frac{4}{3} sum_{nge 3} frac{1}{3^n n} \ &= frac{1}{54} - frac{4}{3} left(frac{1}{3} + frac{1}{9 cdot 2} right) + frac{4}{3} sum_{nge 1} frac{1}{3^n n} \ &= - frac{1}{2} + frac{4}{3} sum_{nge 1} frac{1}{3^n n}. end{align}$$








            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 19 '18 at 19:54

























            answered Dec 19 '18 at 19:46









            angryavianangryavian

            42.2k23481




            42.2k23481












            • $begingroup$
              Can you explain how you get this? $ sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} $? I know that $ frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)} $ but you also take $1/n$ before and it transformed into factors like that: $ frac{1}{3^{n-1}} $
              $endgroup$
              – VirtualUser
              Dec 19 '18 at 20:36












            • $begingroup$
              @VirtualUser Try writing out the first few terms of the sum $sum_{n ge 2} left(frac{1}{3^n(n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)}right)$ explicitly and it should become clear.
              $endgroup$
              – angryavian
              Dec 20 '18 at 16:45










            • $begingroup$
              You have just sorted elements by n'th term, right?
              $endgroup$
              – VirtualUser
              Dec 20 '18 at 18:20


















            • $begingroup$
              Can you explain how you get this? $ sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} $? I know that $ frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)} $ but you also take $1/n$ before and it transformed into factors like that: $ frac{1}{3^{n-1}} $
              $endgroup$
              – VirtualUser
              Dec 19 '18 at 20:36












            • $begingroup$
              @VirtualUser Try writing out the first few terms of the sum $sum_{n ge 2} left(frac{1}{3^n(n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)}right)$ explicitly and it should become clear.
              $endgroup$
              – angryavian
              Dec 20 '18 at 16:45










            • $begingroup$
              You have just sorted elements by n'th term, right?
              $endgroup$
              – VirtualUser
              Dec 20 '18 at 18:20
















            $begingroup$
            Can you explain how you get this? $ sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} $? I know that $ frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)} $ but you also take $1/n$ before and it transformed into factors like that: $ frac{1}{3^{n-1}} $
            $endgroup$
            – VirtualUser
            Dec 19 '18 at 20:36






            $begingroup$
            Can you explain how you get this? $ sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} $? I know that $ frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)} $ but you also take $1/n$ before and it transformed into factors like that: $ frac{1}{3^{n-1}} $
            $endgroup$
            – VirtualUser
            Dec 19 '18 at 20:36














            $begingroup$
            @VirtualUser Try writing out the first few terms of the sum $sum_{n ge 2} left(frac{1}{3^n(n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)}right)$ explicitly and it should become clear.
            $endgroup$
            – angryavian
            Dec 20 '18 at 16:45




            $begingroup$
            @VirtualUser Try writing out the first few terms of the sum $sum_{n ge 2} left(frac{1}{3^n(n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)}right)$ explicitly and it should become clear.
            $endgroup$
            – angryavian
            Dec 20 '18 at 16:45












            $begingroup$
            You have just sorted elements by n'th term, right?
            $endgroup$
            – VirtualUser
            Dec 20 '18 at 18:20




            $begingroup$
            You have just sorted elements by n'th term, right?
            $endgroup$
            – VirtualUser
            Dec 20 '18 at 18:20











            2












            $begingroup$

            hint



            $$frac{2}{n^3-n}=frac{-2}{n}+frac{1}{n-1}+frac{1}{n+1}$$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              hint



              $$frac{2}{n^3-n}=frac{-2}{n}+frac{1}{n-1}+frac{1}{n+1}$$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                hint



                $$frac{2}{n^3-n}=frac{-2}{n}+frac{1}{n-1}+frac{1}{n+1}$$






                share|cite|improve this answer









                $endgroup$



                hint



                $$frac{2}{n^3-n}=frac{-2}{n}+frac{1}{n-1}+frac{1}{n+1}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 19 '18 at 19:40









                hamam_Abdallahhamam_Abdallah

                38.2k21634




                38.2k21634






























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