InterviewSolution
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InterviewSolution
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Let `S_(n),n=1,2,3,"…"` be the sum of infinite geometric series, whose first term is n and the common ratio is `(1)/(n+1)`. Evaluate `lim_(n to oo)(S_(1)S_(n)+S_(2)S_(n-1)+S_(3)S_(n-2)+"..."+S_(n)S_(1))/(S_(1)^(2)+S_(2)^(2)+"......"+S_(n)^(2))`. |
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Answer» `:. S_(n)=(n)/(1-(1)/(n+1)) impliesS_(n)=n+1` `S_(1)S_(n)+S_(2)S_(n-1)+S_(3)S_(n-2)+"..."+S_(n)S_(1)` `=sum_(r=1)^(n)S_(r)S_(n-r-1)=sum_(r=1)^(n)(r+1)(n-r+2)` `=sum_(r=1)^(n)[(n+1)r-r^(2)+(n+2)]` `=(n+1)sumr-sum_(r=1)^(n)r^(2)+(n+2)sum_(r=1)^(n)1` `=(n+1)sumn-sumn^(2)+(n+2)*n` `=((n+1)n(n+1))/(2)-(n(n+1)(2n+1))/(6)+(n+2)n` `(n)/(6)(n^(2)+9n+14)" " "....(i)"` and `S_(1)^(2)+S_(1)^(2)+"...."+S_(n)^(2)=sum_(r=1)^(n)S_(r)^(2)=sum_(r=1)^(n)(r+1)^(2)=sum_(r=0)^(n)(r+1)^(2)-1^(2)` `=((n+1)(n+2)(2n+3))/(6)-1` `(n)/(6)(2n^(2)+9+13)" " ".....(ii)"` From Eqs. (i) and (ii), we get `lim_(n to oo)(S_(1)S_(n)+S_(2)S_(n-1)+S_(3)S_(n-2)+"..."+S_(n)S_(1))/(S_(1)^(2)+S_(2)^(2)+"......"+S_(n)^(2))` `=lim_(n to oo)((n)/(6)(n^(2)+9n+14))/((n)/(6)(2n^(2)+9n+13))=lim_(n to oo)((1+(9)/(n)+(14)/(n^(2))))/((2+(9)/(n)+(13)/(n^(2))))` `(1+0+0)/(2+0+0)=(1)/(2)`. |
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