There's Something About The Numerator

Calculus Level 4

f ( x ) = 1 + x + x 2 + x 3 + ⋯ + x n \large f(x)=1+x+x^2+x^3+\cdots +x^n

For f ( x ) f(x) as defined above, find lim ⁡ n → ∞ ω f ′ ( ω ) ( ω − 1 ) n + 16 \displaystyle \lim_{n\to\infty}\frac{\omega f'(\omega)(\omega -1)}{n+16} , where ω \omega is the primitive ( n + 1 ) th (n+1)^{\text{th}} root of unity .


The answer is 1.

Sravanth C. by Sravanth C. , 21, India

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1 solution

Sravanth C.
Feb 16, 2017

f ′ ( ω ) = n ω n − 1 + ( n − 1 ) ω n − 2 + ⋯ + 2 ω + 1 ( ω − 1 ) f ′ ( ω ) = n ω n + n ω n − 1 − ω n − 2 + n ω n − 3 + n ω n − 2 − 2 ω n − 3 + n ω n − 4 − 3 ω n − 4 + ⋯ + ω − 1 = n ω n − ( ω n − 1 + ω n − 2 + ⋯ + 1 ) = n ω n − ( − ω n ) ( ∵ 1 + ω + ω 2 + ⋯ + ω n = 0 ) = ω n ( n + 1 ) \begin{aligned} f'(\omega)&=n\omega ^{n-1}+(n-1)\omega ^{n-2}+\cdots +2\omega +1\\ (\omega -1)f'(\omega)&=n\omega ^n+n\omega ^{n-1}-\omega ^{n-2}+n\omega ^{n-3}+n\omega ^{n-2}-2\omega ^{n-3}+n\omega ^{n-4}-3\omega ^{n-4}+\cdots +\omega -1\\ &=n\omega ^n-( \omega ^{n-1}+\omega ^{n-2}+\cdots +1)\\ &=n\omega ^n-(- \omega ^n)\quad\quad(\because 1+\omega +\omega ^2+\cdots +\omega ^n =0)\\ &=\omega ^n(n+1) \end{aligned}

Hence the value of the limit would be lim ⁡ n → ∞ ω n + 1 ( n + 1 ) n + 16 = 1 ( ∵ ω n + 1 = 1 ) \lim_{n\to\infty}\frac{\omega ^{n+1}(n+1)}{n+16}=1\quad\quad (\because \omega ^{n+1}=1)

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