Second Symmetric Sum

Algebra Level 4

lim ⁡ n → ∞ S 1 S n + S 2 S n − 1 + S 3 S n − 2 + S 4 S n − 3 + … + S n S 1 S 1 2 + S 2 2 + S 3 2 + … S n 2 \displaystyle \lim_{n \to \infty } \frac {S_1 \ S_n + S_2 \ S_{n-1} + S_3 \ S_{n-2} + S_4 \ S_{n-3} + \ldots + S_n S_1 }{S_1 ^2 + S_2 ^2+S_3 ^2 + \ldots S_n ^2}

For natural numbers n n , denote S n S_n as the sum of an infinite geometric series whose first term is n n and common ratio of 1 n + 1 \frac 1 {n+1}

If the limit above equals to X X , what is the value of 10 × X 10 \times X ?


The answer is 5.

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Rahul Chandani by Rahul Chandani , 23, India

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1 solution

Rahul Chandani
Nov 22, 2014

S n = n 1 − 1 n + 1 S n = n + 1 ∴ S 1 . S n + S 2 S n − 1 + S 3 S n − 3 + . . . . . . . . + S n S 1 = 2 ( n + 1 ) + 3 ( n ) + 4 ( n − 1 ) + . . . . . . + ( n + 1 ) ( 2 ) = ∑ r = 1 n ( r + 1 ) ( n − r + 2 ) = ∑ r = 1 n ( n r − r 2 + 2 r + n − r + 2 ) = ∑ r = 1 n { ( n + 1 ) r − r 2 + ( n + 2 ) } = ( n + 1 ) ∑ r = 1 n r − ∑ r = 1 n r 2 + ( n + 2 ) ∑ r = 1 n 1 = ( n + 1 ) n ( n + 1 ) 2 − n ( n + 1 ) ( 2 n + 1 ) 6 + n ( n + 2 ) = n 6 { n 2 + 9 n + 14 } . . . . . ( 1 ) a n d S 1 2 + S 2 2 + S 3 2 + S 4 2 + . . . . . . . . + S n 2 = n 6 { 2 n 2 + 9 n + 13 } . . . . ( 2 ) F r o m ( 1 ) a n d ( 2 ) , w e g e t , S 1 . S n + S 2 S n − 1 + S 3 S n − 3 + . . . . . . . . + S n S 1 S 1 2 + S 2 2 + S 3 2 + S 4 2 + . . . . . . . . + S n 2 = n 2 + 9 n + 14 2 n 2 + 9 n + 13 = 1 + 9 n + 14 n 2 2 + 9 n + 13 n 2 N o w , lim ⁡ n → ∞ S 1 . S n + S 2 S n − 1 + S 3 S n − 3 + . . . . . . . . + S n S 1 S 1 2 + S 2 2 + S 3 2 + S 4 2 + . . . . . . . . + S n 2 = lim ⁡ n → ∞ 1 + 9 n + 14 n 2 2 + 9 n + 13 n 2 = 1 + 0 + 0 2 + 0 + 0 = 1 2 N o w X = 1 2 ∴ 10 X = 5 { S }_{ n }\quad =\quad \frac { n }{ 1\quad -\quad \frac { 1 }{ n+1 } } \\ { S }_{ n }\quad =\quad n+1\\ \therefore \quad \quad { S }_{ 1 }.{ S }_{ n }+{ S }_{ 2 }{ S }_{ n-1 }+{ S }_{ 3 }{ S }_{ n-3 }+........+{ S }_{ n }{ S }_{ 1 }\quad \\ =\quad \quad \quad \quad 2(n+1)\quad +\quad 3(n)\quad +\quad 4(n-1)\quad +\quad ......\quad +\quad (n+1)(2)\\ =\quad \quad \quad \quad \sum _{ r=1 }^{ n }{ (r+1)(n-r+2) } \\ =\quad \quad \quad \quad \sum _{ r=1 }^{ n }{ (nr-{ r }^{ 2 } } +2r+n-r+2)\\ =\quad \quad \quad \quad \sum _{ r=1 }^{ n }{ \left\{ (n+1)r\quad -\quad { r }^{ 2 }\quad +\quad (n+2) \right\} } \\ =\quad \quad \quad \quad (n+1)\sum _{ r=1 }^{ n }{ r } \quad -\quad \sum _{ r=1 }^{ n }{ { r }^{ 2 } } \quad +\quad (n+2)\sum _{ r=1 }^{ n }{ 1 } \\ =\quad \quad \quad \quad \frac { (n+1)n(n+1) }{ 2 } \quad -\quad \frac { n(n+1)(2n+1) }{ 6 } \quad +\quad n(n+2)\\ =\quad \quad \quad \quad \frac { n }{ 6 } \left\{ { n }^{ 2 }+9n+14 \right\} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad .....(1)\\ \\ and\quad \quad \quad { S }_{ 1 }^{ 2 }+{ S }_{ 2 }^{ 2 }+{ S }_{ 3 }^{ 2 }+{ S }_{ 4 }^{ 2 }+\quad ........+{ S }_{ n }^{ 2 }\\ =\quad \quad \quad \quad \quad \frac { n }{ 6 } \left\{ 2{ n }^{ 2 }+9n+13 \right\} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ....(2)\\ From\quad (1)\quad and\quad (2),\quad we\quad get,\\ \\ \frac { { S }_{ 1 }.{ S }_{ n }+{ S }_{ 2 }{ S }_{ n-1 }+{ S }_{ 3 }{ S }_{ n-3 }+........+{ S }_{ n }{ S }_{ 1 } }{ { S }_{ 1 }^{ 2 }+{ S }_{ 2 }^{ 2 }+{ S }_{ 3 }^{ 2 }+{ S }_{ 4 }^{ 2 }+\quad ........+{ S }_{ n }^{ 2 } } =\quad \frac { { n }^{ 2 }+9n+14 }{ 2{ n }^{ 2 }+9n+13 } \\ =\quad \quad \quad \quad \frac { 1\quad +\quad \frac { 9 }{ n } +\quad \frac { 14 }{ { n }^{ 2 } } }{ 2\quad +\quad \frac { 9 }{ n } \quad +\frac { 13 }{ { n }^{ 2 } } } \\ \\ Now,\quad \lim _{ n\rightarrow \infty }{ \frac { { S }_{ 1 }.{ S }_{ n }+{ S }_{ 2 }{ S }_{ n-1 }+{ S }_{ 3 }{ S }_{ n-3 }+........+{ S }_{ n }{ S }_{ 1 } }{ { S }_{ 1 }^{ 2 }+{ S }_{ 2 }^{ 2 }+{ S }_{ 3 }^{ 2 }+{ S }_{ 4 }^{ 2 }+\quad ........+{ S }_{ n }^{ 2 } } } \\ =\quad \quad \quad \quad \lim _{ n\rightarrow \infty }{ \frac { 1\quad +\quad \frac { 9 }{ n } +\quad \frac { 14 }{ { n }^{ 2 } } }{ 2\quad +\quad \frac { 9 }{ n } \quad +\frac { 13 }{ { n }^{ 2 } } } } \\ =\quad \quad \quad \quad \quad \frac { 1+0+0 }{ 2+0+0 } =\quad \frac { 1 }{ 2 } \\ Now\quad X \quad =\quad \frac { 1 }{ 2 } \\ \therefore \quad 10X \quad =\quad 5

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For your Latex to display, you have to place it within the brackets: \ ( Latex code \ ) \backslash ( \text{Latex code } \backslash ) . I've edited your solution for your reference.

Calvin Lin Staff - 6 years, 6 months ago

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Thanku! :)

Rahul Chandani - 6 years, 6 months ago

Great algebraic solution, but one could easily apply L'Hopital's rule to evaluate the limit in the second half. For some reason, I thought the question asked for 10 b a 10\frac{b}{a} .

Jake Lai - 6 years, 6 months ago

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