Bounding the nth term

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Bounding the nth term

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Math-Problem-Solving

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Math-Problem-Solving

#1 h Apr 1, 2025, 5:32 AM

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We define a sequence $(a_n)_{n\ge 0}$ by ${a_0}=1+ \frac{1}{2k}$ where $k \in N$ and ${a_n} = {{a^2}_{n-1}} - {a_{n-1}}+1 for all {n \ge 1}.$ Then prove that $\frac{1}{k} - \frac{1}{k^2} +1 < a_k < 1+ \frac{1}{k}.$

This post has been edited 2 times. Last edited by Math-Problem-Solving, Apr 1, 2025, 11:52 AM

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pco

#2 h Apr 1, 2025, 6:58 AM

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Math-Problem-Solving wrote:

We define a sequence $(a_n)_{n\ge 0}$ by ${a_1}=1+ \frac{1}{2k}$ where $k \in N$ and ${a_n} = {{a^2}_{n-1}} - {a_{n-1}}+1 for all {n \ge 1}.$ Then prove that $\frac{1}{n} - \frac{1}{n^2} +1 < a_n < 1+ \frac{1}{n}.$

Wrong.
Choose $k=1$ and you get $a_1=\frac 32$ and $a_2=\frac 74>1+\frac 12=\frac 32$

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Math-Problem-Solving

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Math-Problem-Solving

#3 h Apr 1, 2025, 8:52 AM

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pco wrote:

Math-Problem-Solving wrote:

We define a sequence $(a_n)_{n\ge 0}$ by ${a_1}=1+ \frac{1}{2k}$ where $k \in N$ and ${a_n} = {{a^2}_{n-1}} - {a_{n-1}}+1 for all {n \ge 1}.$ Then prove that $\frac{1}{n} - \frac{1}{n^2} +1 < a_n < 1+ \frac{1}{n}.$

Wrong.
Choose $k=1$ and you get $a_1=\frac 32$ and $a_2=\frac 74>1+\frac 12=\frac 32$

Fixed! That would be ${a_0}=1+ \frac{1}{2k}$ , not ${a_1}=1+ \frac{1}{2k}$ .

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pco

#4 h Apr 1, 2025, 9:32 AM

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Math-Problem-Solving wrote:

We define a sequence $(a_n)_{n\ge 0}$ by ${a_0}=1+ \frac{1}{2k}$ where $k \in N$ and ${a_n} = {{a^2}_{n-1}} - {a_{n-1}}+1 for all {n \ge 1}.$ Then prove that $\frac{1}{n} - \frac{1}{n^2} +1 < a_n < 1+ \frac{1}{n}.$

Still wrong.
Choose $k=1$ and you get $a_0=\frac 32$ , $a_1=\frac 74$ and $a_2=\frac{37}{16}>1+\frac 12$

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Math-Problem-Solving

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Math-Problem-Solving

#5 h Apr 1, 2025, 11:53 AM

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Probably it is correct now.

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Math-Problem-Solving

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Math-Problem-Solving

#6 h Apr 1, 2025, 3:35 PM

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Try to prove it.

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