Sum of series $sqrt{1+frac1{n^2}+frac1{(n+1){}^2}}$












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How to solve this sum?$$sqrt{1+frac1{1^2}+frac1{2^2}}+sqrt{1+frac1{2^2}+frac1{3^2}}+cdots+sqrt{1+frac1{19^2}+frac1{20^2}}$$
I assumed it to be a sum of $sqrt{1+dfrac{1}{n^2} + dfrac{1}{(n+1)^2}}$ but It complicates the powers and I am unable to factorise further.






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    $begingroup$


    How to solve this sum?$$sqrt{1+frac1{1^2}+frac1{2^2}}+sqrt{1+frac1{2^2}+frac1{3^2}}+cdots+sqrt{1+frac1{19^2}+frac1{20^2}}$$
    I assumed it to be a sum of $sqrt{1+dfrac{1}{n^2} + dfrac{1}{(n+1)^2}}$ but It complicates the powers and I am unable to factorise further.






    sequences-and-series







    share|cite|improve this question











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      $begingroup$


      How to solve this sum?$$sqrt{1+frac1{1^2}+frac1{2^2}}+sqrt{1+frac1{2^2}+frac1{3^2}}+cdots+sqrt{1+frac1{19^2}+frac1{20^2}}$$
      I assumed it to be a sum of $sqrt{1+dfrac{1}{n^2} + dfrac{1}{(n+1)^2}}$ but It complicates the powers and I am unable to factorise further.






      sequences-and-series







      share|cite|improve this question











      $endgroup$




      How to solve this sum?$$sqrt{1+frac1{1^2}+frac1{2^2}}+sqrt{1+frac1{2^2}+frac1{3^2}}+cdots+sqrt{1+frac1{19^2}+frac1{20^2}}$$
      I assumed it to be a sum of $sqrt{1+dfrac{1}{n^2} + dfrac{1}{(n+1)^2}}$ but It complicates the powers and I am unable to factorise further.






      sequences-and-series




      sequences-and-series






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      edited Jan 3 at 12:26









      Saad

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      asked Jan 3 at 12:18









      Ice InkberryIce Inkberry

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          $begingroup$

          $$1+dfrac1{n^2}+dfrac1{(n+1)^2}=dfrac{n^2+(n+1)^2+(n^2+n)^2}{n^2(n+1)^2}=dfrac{n^4+2n^3+3n^2+2n+1}{(n^2+n)^2}$$



          $n^4+2n^3+3n^2+2n+1=(n^2)^2+2n^2cdot1+2ncdot1+(n)^2+1^2+2n^2cdot n=(n^2+n+1)^2$



          Now $$dfrac{n^2+n+1}{n(n+1)}=1+dfrac1{n(n+1)}=1+dfrac{n+1-n}{n(n+1)}=?$$



          See also: Telescoping series






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          • 1




            $begingroup$
            $$left(1+dfrac1n-dfrac1{n+1}right)^2=1+dfrac1{n^2}+dfrac1{(n+1)^2}+dfrac2n-dfrac2{n+1}-dfrac2{n(n+1)}$$ $dfrac1n-dfrac1{n+1}-dfrac1{n(n+1)}=dfrac{n+1-n-1}{n(n+1)}=0$
            $endgroup$
            – lab bhattacharjee
            Jan 3 at 12:31






          • 1




            $begingroup$
            $$left(dfrac1a+dfrac1b+dfrac1cright)^2=dfrac1{a^2}+dfrac1{b^2}+dfrac1{c^2}$$ will hold if $$a+b+c=0$$ Here $a=1,b=n,c=?$
            $endgroup$
            – lab bhattacharjee
            Jan 3 at 12:44













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          $begingroup$

          $$1+dfrac1{n^2}+dfrac1{(n+1)^2}=dfrac{n^2+(n+1)^2+(n^2+n)^2}{n^2(n+1)^2}=dfrac{n^4+2n^3+3n^2+2n+1}{(n^2+n)^2}$$



          $n^4+2n^3+3n^2+2n+1=(n^2)^2+2n^2cdot1+2ncdot1+(n)^2+1^2+2n^2cdot n=(n^2+n+1)^2$



          Now $$dfrac{n^2+n+1}{n(n+1)}=1+dfrac1{n(n+1)}=1+dfrac{n+1-n}{n(n+1)}=?$$



          See also: Telescoping series






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            $$left(1+dfrac1n-dfrac1{n+1}right)^2=1+dfrac1{n^2}+dfrac1{(n+1)^2}+dfrac2n-dfrac2{n+1}-dfrac2{n(n+1)}$$ $dfrac1n-dfrac1{n+1}-dfrac1{n(n+1)}=dfrac{n+1-n-1}{n(n+1)}=0$
            $endgroup$
            – lab bhattacharjee
            Jan 3 at 12:31






          • 1




            $begingroup$
            $$left(dfrac1a+dfrac1b+dfrac1cright)^2=dfrac1{a^2}+dfrac1{b^2}+dfrac1{c^2}$$ will hold if $$a+b+c=0$$ Here $a=1,b=n,c=?$
            $endgroup$
            – lab bhattacharjee
            Jan 3 at 12:44


















          2












          $begingroup$

          $$1+dfrac1{n^2}+dfrac1{(n+1)^2}=dfrac{n^2+(n+1)^2+(n^2+n)^2}{n^2(n+1)^2}=dfrac{n^4+2n^3+3n^2+2n+1}{(n^2+n)^2}$$



          $n^4+2n^3+3n^2+2n+1=(n^2)^2+2n^2cdot1+2ncdot1+(n)^2+1^2+2n^2cdot n=(n^2+n+1)^2$



          Now $$dfrac{n^2+n+1}{n(n+1)}=1+dfrac1{n(n+1)}=1+dfrac{n+1-n}{n(n+1)}=?$$



          See also: Telescoping series






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            $$left(1+dfrac1n-dfrac1{n+1}right)^2=1+dfrac1{n^2}+dfrac1{(n+1)^2}+dfrac2n-dfrac2{n+1}-dfrac2{n(n+1)}$$ $dfrac1n-dfrac1{n+1}-dfrac1{n(n+1)}=dfrac{n+1-n-1}{n(n+1)}=0$
            $endgroup$
            – lab bhattacharjee
            Jan 3 at 12:31






          • 1




            $begingroup$
            $$left(dfrac1a+dfrac1b+dfrac1cright)^2=dfrac1{a^2}+dfrac1{b^2}+dfrac1{c^2}$$ will hold if $$a+b+c=0$$ Here $a=1,b=n,c=?$
            $endgroup$
            – lab bhattacharjee
            Jan 3 at 12:44
















          2












          2








          2





          $begingroup$

          $$1+dfrac1{n^2}+dfrac1{(n+1)^2}=dfrac{n^2+(n+1)^2+(n^2+n)^2}{n^2(n+1)^2}=dfrac{n^4+2n^3+3n^2+2n+1}{(n^2+n)^2}$$



          $n^4+2n^3+3n^2+2n+1=(n^2)^2+2n^2cdot1+2ncdot1+(n)^2+1^2+2n^2cdot n=(n^2+n+1)^2$



          Now $$dfrac{n^2+n+1}{n(n+1)}=1+dfrac1{n(n+1)}=1+dfrac{n+1-n}{n(n+1)}=?$$



          See also: Telescoping series






          share|cite|improve this answer









          $endgroup$



          $$1+dfrac1{n^2}+dfrac1{(n+1)^2}=dfrac{n^2+(n+1)^2+(n^2+n)^2}{n^2(n+1)^2}=dfrac{n^4+2n^3+3n^2+2n+1}{(n^2+n)^2}$$



          $n^4+2n^3+3n^2+2n+1=(n^2)^2+2n^2cdot1+2ncdot1+(n)^2+1^2+2n^2cdot n=(n^2+n+1)^2$



          Now $$dfrac{n^2+n+1}{n(n+1)}=1+dfrac1{n(n+1)}=1+dfrac{n+1-n}{n(n+1)}=?$$



          See also: Telescoping series







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 3 at 12:25









          lab bhattacharjeelab bhattacharjee

          225k15157275




          225k15157275








          • 1




            $begingroup$
            $$left(1+dfrac1n-dfrac1{n+1}right)^2=1+dfrac1{n^2}+dfrac1{(n+1)^2}+dfrac2n-dfrac2{n+1}-dfrac2{n(n+1)}$$ $dfrac1n-dfrac1{n+1}-dfrac1{n(n+1)}=dfrac{n+1-n-1}{n(n+1)}=0$
            $endgroup$
            – lab bhattacharjee
            Jan 3 at 12:31






          • 1




            $begingroup$
            $$left(dfrac1a+dfrac1b+dfrac1cright)^2=dfrac1{a^2}+dfrac1{b^2}+dfrac1{c^2}$$ will hold if $$a+b+c=0$$ Here $a=1,b=n,c=?$
            $endgroup$
            – lab bhattacharjee
            Jan 3 at 12:44
















          • 1




            $begingroup$
            $$left(1+dfrac1n-dfrac1{n+1}right)^2=1+dfrac1{n^2}+dfrac1{(n+1)^2}+dfrac2n-dfrac2{n+1}-dfrac2{n(n+1)}$$ $dfrac1n-dfrac1{n+1}-dfrac1{n(n+1)}=dfrac{n+1-n-1}{n(n+1)}=0$
            $endgroup$
            – lab bhattacharjee
            Jan 3 at 12:31






          • 1




            $begingroup$
            $$left(dfrac1a+dfrac1b+dfrac1cright)^2=dfrac1{a^2}+dfrac1{b^2}+dfrac1{c^2}$$ will hold if $$a+b+c=0$$ Here $a=1,b=n,c=?$
            $endgroup$
            – lab bhattacharjee
            Jan 3 at 12:44










          1




          1




          $begingroup$
          $$left(1+dfrac1n-dfrac1{n+1}right)^2=1+dfrac1{n^2}+dfrac1{(n+1)^2}+dfrac2n-dfrac2{n+1}-dfrac2{n(n+1)}$$ $dfrac1n-dfrac1{n+1}-dfrac1{n(n+1)}=dfrac{n+1-n-1}{n(n+1)}=0$
          $endgroup$
          – lab bhattacharjee
          Jan 3 at 12:31




          $begingroup$
          $$left(1+dfrac1n-dfrac1{n+1}right)^2=1+dfrac1{n^2}+dfrac1{(n+1)^2}+dfrac2n-dfrac2{n+1}-dfrac2{n(n+1)}$$ $dfrac1n-dfrac1{n+1}-dfrac1{n(n+1)}=dfrac{n+1-n-1}{n(n+1)}=0$
          $endgroup$
          – lab bhattacharjee
          Jan 3 at 12:31




          1




          1




          $begingroup$
          $$left(dfrac1a+dfrac1b+dfrac1cright)^2=dfrac1{a^2}+dfrac1{b^2}+dfrac1{c^2}$$ will hold if $$a+b+c=0$$ Here $a=1,b=n,c=?$
          $endgroup$
          – lab bhattacharjee
          Jan 3 at 12:44






          $begingroup$
          $$left(dfrac1a+dfrac1b+dfrac1cright)^2=dfrac1{a^2}+dfrac1{b^2}+dfrac1{c^2}$$ will hold if $$a+b+c=0$$ Here $a=1,b=n,c=?$
          $endgroup$
          – lab bhattacharjee
          Jan 3 at 12:44




















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