What is ∑ r = 1 n r 7 \sum _{r=1}^n r^7 ?

Calculus Level 4

Let M n = ∑ r = 1 n r 4 , N n = ∑ r = 1 n r 2 \displaystyle M_n= \sum_{r=1}^n r^4 , N_n= \sum_{r=1}^n r^2 and Y n = ∑ r = 1 n r 7 \displaystyle Y_n= \sum_{r=1}^n r^7 . If lim ⁡ n → ∞ M n N n Y n \displaystyle \lim_{n\to\infty} \dfrac{M_n N_n}{Y_n} can be expressed as A B \dfrac AB , where A A and B B are coprime positive integers.

Find A + B A+B .

You may want to read the wiki page: Method of undetermined coefficients .


The answer is 23.

Rishabh Deep Singh by Rishabh Deep Singh , 23, India

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1 solution

∣ M e t h o d 1 ∣ \left| Method\quad 1 \right|

lim ⁡ n → ∞ ( ∑ r = 1 n r 4 ) ( ∑ r = 1 n r 2 ) ( ∑ r = 1 n r 7 ) \lim _{ n\rightarrow \infty }{ \frac { \left( \sum _{ r=1 }^{ n }{ { r }^{ 4 } } \right) \left( \sum _{ r=1 }^{ n }{ { r }^{ 2 } } \right) }{ \left( \sum _{ r=1 }^{ n }{ { r }^{ 7 } } \right) } }

= lim ⁡ n → ∞ ( ∑ r = 1 n ( r n ) 4 ) ( ∑ r = 1 n ( r n ) 2 ) ( ∑ r = 1 n ( r n ) 7 ) n 6 n 7 =\lim _{ n\rightarrow \infty }{ \frac { \left( \sum _{ r=1 }^{ n }{ { \left( \frac { r }{ n } \right) }^{ 4 } } \right) \left( \sum _{ r=1 }^{ n }{ { \left( \frac { r }{ n } \right) }^{ 2 } } \right) }{ \left( \sum _{ r=1 }^{ n }{ { \left( \frac { r }{ n } \right) }^{ 7 } } \right) } } \frac { { n }^{ 6 } }{ { n }^{ 7 } }

= lim ⁡ n → ∞ 1 n ( ∑ r = 1 n ( r n ) 4 ) ( ∑ r = 1 n ( r n ) 2 ) ( ∑ r = 1 n ( r n ) 7 ) =\lim _{ n\rightarrow \infty }{ \frac { 1 }{ n } \frac { \left( \sum _{ r=1 }^{ n }{ { \left( \frac { r }{ n } \right) }^{ 4 } } \right) \left( \sum _{ r=1 }^{ n }{ { \left( \frac { r }{ n } \right) }^{ 2 } } \right) }{ \left( \sum _{ r=1 }^{ n }{ { \left( \frac { r }{ n } \right) }^{ 7 } } \right) } }

= ( ∫ 0 1 x 4 d x ) ( ∫ 0 1 x 2 d x ) ( ∫ 0 1 x 7 d x ) =\frac { \left( \int _{ 0 }^{ 1 }{ { x }^{ 4 }dx } \right) \left( \int _{ 0 }^{ 1 }{ { x }^{ 2 }dx } \right) }{ \left( \int _{ 0 }^{ 1 }{ { x }^{ 7 }dx } \right) }

= ( 1 5 ) ( 1 3 ) ( 1 8 ) = 8 15 = A B =\frac { \left( \frac { 1 }{ 5 } \right) \left( \frac { 1 }{ 3 } \right) }{ \left( \frac { 1 }{ 8 } \right) } =\frac { 8 }{ 15 } =\frac { A }{ B }

Now A+B=23

∣ M e t h o d 2 ∣ \left| Method\quad 2 \right|

∑ r = 1 n r n \sum _{ r=1 }^{ n }{ { r }^{ n } } Is a Polynomial With Leading Coefficient 1 n + 1 \frac { 1 }{ n+1 } And Of Degree n + 1 n+1

lim ⁡ n → ∞ ( ∑ r = 1 n r 4 ) ( ∑ r = 1 n r 2 ) ( ∑ r = 1 n r 7 ) \lim _{ n\rightarrow \infty }{ \frac { \left( \sum _{ r=1 }^{ n }{ { r }^{ 4 } } \right) \left( \sum _{ r=1 }^{ n }{ { r }^{ 2 } } \right) }{ \left( \sum _{ r=1 }^{ n }{ { r }^{ 7 } } \right) } }

= lim ⁡ n → ∞ ( n 5 5 + f ( x ) ) ( n 2 2 + g ( x ) ) ( n 8 8 + h ( x ) ) =\lim _{ n\rightarrow \infty }{ \frac { \left( \frac { { n }^{ 5 } }{ 5 } +f\left( x \right) \right) \left( \frac { { n }^{ 2 } }{ 2 } +g\left( x \right) \right) }{ \left( \frac { { n }^{ 8 } }{ 8 } +h\left( x \right) \right) } }

f ( x ) I s a P o l y n o m i a l O f d e g r e e 4 g ( x ) I s a P o l y n o m i a l O f d e g r e e 2 h ( x ) I s a P o l y n o m i a l O f d e g r e e 7 f\left( x \right) \quad Is\quad a\quad Polynomial\quad Of\quad degree\quad 4\\ g\left( x \right) \quad Is\quad a\quad Polynomial\quad Of\quad degree\quad 2\\ h\left( x \right) \quad Is\quad a\quad Polynomial\quad Of\quad degree\quad 7\\

= lim ⁡ n → ∞ ( 1 5 + f ( x ) n 5 ) ( 1 3 + g ( x ) n 3 ) ( 1 8 + h ( x ) n 8 ) = ( 1 5 ) ( 1 3 ) ( 1 8 ) = 8 15 =\lim _{ n\rightarrow \infty }{ \frac { \left( \frac { 1 }{ 5 } +\frac { f\left( x \right) }{ { n }^{ 5 } } \right) \left( \frac { 1 }{ 3 } +\frac { g\left( x \right) }{ { n }^{ 3 } } \right) }{ \left( \frac { 1 }{ 8 } +\frac { h\left( x \right) }{ { n }^{ 8 } } \right) } } \\ =\frac { \left( \frac { 1 }{ 5 } \right) \left( \frac { 1 }{ 3 } \right) }{ \left( \frac { 1 }{ 8 } \right) } =\frac { 8 }{ 15 }

I hope You guys Like my Approach.

5 Helpful 0 Interesting 1 Brilliant 0 Confused

Moderator note:

Good first approach.

For method 2, you should explain why " ∑ \sum is a polynomial with leading coefficient 1 n + 1 \frac{1}{n+1} ". At least mention that this is well-known / can be calculated in the following way / state the summation explicitly.

Note: In general, avoid making people jump through hoops in order to enter the answer. IE After finding that A + B = 23 A+B = 23 , they shouldn't then be asked to run a mile, touch their finger to their noise, find which prime number it is, square the value, multiply it by 2, etc.

I've udpated the answer to 23.

Calvin Lin Staff - 5 years ago

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