Summing Deltas

Calculus Level 5

lim n 1 n 2 k = 1 n S k \large \displaystyle \underset { n\rightarrow \infty }{ \lim } \frac { 1 }{ { n }^{ 2 } } \sum _{ k=1 }^{ n }{ { S }_{ k } }

Let S k { S }_{ k } denote area of triangle A O B k { AOB }_{ k } with 2 given sides of O A = 1 OA=1 , O B k = k {OB }_{ k }= k and A O B k = k π 2 n \angle {AOB }_{ k }=\dfrac { k\pi }{ 2n } for positive integer k k .

If the value of the limit above can be expressed as C × π D C\times \pi^D , where C C and D D are integers, find the value of C × D C\times D .


The answer is -4.

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1 solution

Anupam Khandelwal
Mar 31, 2015

A r e a o f Δ A O B K = 1 2 . 1. K . sin K π 2 n T h i s i m p l i e s S K = K 2 sin K π 2 n l i m n 1 n 2 K = 1 n K 2 sin K π 2 n l i m n 1 n K = 1 n 1 2 . K n sin ( π 2 . K n ) N o w t h e s u m m a t i o n c a n b e c o n v e r t e d i n t o d e f i n i t e i n t e g r a l . L e t K n = x 1 n d x L o w e r l i m i t l i m K = 1 n K n = 0 U p p e r l i m i t l i m K n n K n = 1 0 1 x 2 sin π x 2 . d x U s i n g I n t e g r a t i o n B y P a r t s w e g e t 1 2 [ x cos π x 2 π 2 ] 0 1 + 1 2 [ sin π x 2 ( π 2 ) 2 ] 0 1 = 2 π 2 H e n c e , C = 2 , D = 2 , C × D = 4. Area\quad of\quad \Delta { AOB }_{ K }\quad =\quad \frac { 1 }{ 2 } .1.K.\sin { \frac { K\pi }{ 2n } } \\ This\quad implies\quad { S }_{ K }\quad =\quad \frac { K }{ 2 } \sin { \frac { K\pi }{ 2n } } \\ \underset { n\rightarrow \infty }{ lim } \quad \frac { 1 }{ { n }^{ 2 } } \sum _{ K=1 }^{ n }{ \frac { K }{ 2 } } \sin { \frac { K\pi }{ 2n } } \\ \underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \sum _{ K=1 }^{ n }{ \frac { 1 }{ 2 } .\frac { K }{ n } \sin { (\frac { \pi }{ 2 } .\frac { K }{ n } ) } } \\ \\ Now\quad the\quad summation\quad can\quad be\quad \\ converted\quad into\quad definite\quad integral.\\ Let\quad \frac { K }{ n } =\quad x\quad \Rightarrow \frac { 1 }{ n } \rightarrow dx\\ Lower\quad limit\quad \rightarrow \underset { K=1\\ n\rightarrow \infty }{ lim } \frac { K }{ n } =0\\ Upper\quad limit\quad \rightarrow \underset { K\rightarrow n\\ n\rightarrow \infty }{ lim } \frac { K }{ n } =1\\ \int _{ 0 }^{ 1 }{ \frac { x }{ 2 } \sin { \frac { \pi x }{ 2 } .dx } } \\ Using\quad Integration\quad By\quad Parts\quad we\quad get\\ \frac { 1 }{ 2 } { \left[ \frac { -x\cos { \frac { \pi x }{ 2 } } }{ \frac { \pi }{ 2 } } \right] }_{ 0 }^{ 1 }\quad +\frac { 1 }{ 2 } { \left[ \frac { \sin { \frac { \pi x }{ 2 } } }{ { \left( \frac { \pi }{ 2 } \right) }^{ 2 } } \right] }_{ 0 }^{ 1 }=\frac { 2 }{ { \pi }^{ 2 } } \\ Hence,\quad C = 2, D = -2, \quad \quad C\times D = -4.

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