Recurring is recurring is recurring is

Algebra Level 4

Let f 1 n ( x ) = x x n + 1 n f_1^n (x) = \dfrac x{\sqrt[n]{x^n+1}} and f q p ( x ) = f ( f ( f ( … ( f ( x ) … ) ⏟ q number of times f^p _q (x) = \underbrace{f(f(f(\ldots (f(x)\ldots)}_{q \text{number of times}} where f 1 p ( x ) = x x p + 1 p f^p _1 (x) = \dfrac x{\sqrt[p]{x^p+1}} .

Evaluate lim ⁡ x → ∞ ∑ r = 1 n ( f r r ( x ) ) r \displaystyle \lim_{x\to\infty} \sum_{r=1}^n (f^r _r (x))^r .

The problem is original. Inspiration .
n ∑ r = 1 n 1 r n\sum _{ r=1 }^{ n }{ \frac { 1 }{ r } } 2 ∑ r = 1 n 1 r 2 2\sum _{ r=1 }^{ n }{ \frac { 1 }{ { r }^{ 2 } } } ∑ r = 1 n 1 r \sum _{ r=1 }^{ n }{ \frac { 1 }{ r } } n ∑ r = 1 n 1 r n n\sum _{ r=1 }^{ n }{ \frac { 1 }{ { r }^{ n } } }

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Rohit Ner by Rohit Ner , 22, India

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1 solution

f 1 r ( x ) = x r x r + 1 r f 2 r ( x ) = x r 2 x r + 1 r = > ( f r r ( x ) ) r = x r r x r + 1 lim ⁡ x → ∞ ∑ r = 1 n ( f r r ( x ) ) r = lim ⁡ x → ∞ ∑ r = 1 n x r r x r + 1 lim ⁡ x → ∞ ∑ r = 1 n 1 r + 1 x r = ∑ r = 1 n 1 r f_{ 1 }^{ r }\left( x \right) =\sqrt [ r ]{ \frac { { x }^{ r } }{ { x }^{ r }+1 } } \\ f_{ 2 }^{ r }\left( x \right) =\sqrt [ r ]{ \frac { { x }^{ r } }{ { 2x }^{ r }+1 } } \\ =>\quad { \left( f_{ r }^{ r }\left( x \right) \right) }^{ r }=\frac { { x }^{ r } }{ r{ x }^{ r }+1 } \\ \lim _{ x\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ { \left( f_{ r }^{ r }\left( x \right) \right) }^{ r }\quad } } =\lim _{ x\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \frac { { x }^{ r } }{ r{ x }^{ r }+1 } } } \\ \lim _{ x\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \frac { 1 }{ r+\frac { 1 }{ { x }^{ r } } } } } =\sum _{ r=1 }^{ n }{ \frac { 1 }{ r } } \\

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Moderator note:

  1. You should demonstrate how the 2nd line is calculated.

  2. You haven't provided any reasoning for why the implication sign is true. Looking at the first 2 base cases tells us nothing. At the very least, state "By induction, we can show that ..."

  3. Be careful when interchanging limits with summation signs. This needs to be justified in the final step.

  1. You should demonstrate how the 2nd line is calculated.

  • You haven't provided any reasoning for why the implication sign is true. Looking at the first 2 base cases tells us nothing. At the very least, state "By induction, we can show that ..."

  • Be careful when interchanging limits with summation signs. This needs to be justified in the final step.

    • Calvin Lin Staff - 5 years, 4 months ago

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