NEPAL TST DAY-2 PROBLEM 1

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NEPAL TST DAY-2 PROBLEM 1

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#1 h Apr 12, 2025, 8:34 AM

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Let the sequence $\{a_n\}_{n \geq 1}$ be defined by
$a_1 = 1, \quad a_{n+1} = a_n + \frac{1}{\sqrt[2024]{a_n}} \quad \text{for } n \geq 1, \, n \in \mathbb{N}$ Prove that
$a_n^{2025} >n^{2024}$ for all positive integers $n \geq 2$ .

$\textbf{Proposed by Prajit Adhikari, Nepal.}$

This post has been edited 1 time. Last edited by Tony_stark0094, Apr 13, 2025, 12:36 AM
Reason: typo

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#3 h Apr 12, 2025, 11:26 AM

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Note that $x + \frac{1}{\sqrt[2024]{x}}$ is an increasing function for $x \geq 1$ by differentiating. Also note that by induction $a_n \geq 1$ for all $n$ .
We will prove the main result by induction, the base case $n=2$ is true, assume true for $n = k \geq 2$ note that $a_{k+1} = a_k + \frac{1}{\sqrt[2024]{a_k}} > k^{2024/2025} + \frac{1}{\sqrt[2025]{k}} = \frac{k+1}{\sqrt[2025]{k}} > (k+1)^{2024/2025}$

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#5 h Apr 12, 2025, 12:23 PM

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@above

ThatApollo777 wrote:

$a_k + \frac{1}{\sqrt[2024]{a_k}} > k^{2024/2025} + \frac{1}{\sqrt[2025]{k}}$

How?

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#6 h Apr 12, 2025, 12:23 PM

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ThatApollo777 wrote:

Note that $x + \frac{1}{\sqrt[2024]{x}}$ is an increasing function for $x \geq 1$ by differentiating. Also note that by induction $a_n \geq 1$ for all $n$ .
We will prove the main result by induction, the base case $n=2$ is true, assume true for $n = k \geq 2$ note that $a_{k+1} = a_k + \frac{1}{\sqrt[2024]{a_k}} > k^{2024/2025} + \frac{1}{\sqrt[2025]{k}} = \frac{k+1}{\sqrt[2025]{k}} > (k+1)^{2024/2025}$

nice sol . I got the same proof

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#8 h Apr 12, 2025, 12:27 PM

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Thapakazi wrote:

@above

ThatApollo777 wrote:

$a_k + \frac{1}{\sqrt[2024]{a_k}} > k^{2024/2025} + \frac{1}{\sqrt[2025]{k}}$

How?

$a_k+\frac{1}{\sqrt[2024]{a_k}}$ is a increasing function and $a_k > k^{\frac{2024}{2025}}$
$\implies a_k+\frac{1}{\sqrt[2024]{a_k}} >k^{\frac{2024}{2025}}+\frac{1}{\sqrt[2025]{k}}=\frac{k+1}{\sqrt[2025]{k}}$
and $\frac{1}{\sqrt[2025]k}>\frac{1}{\sqrt[2025]{k+1}} \implies \frac{k+1}{\sqrt[2025]k}>(k+1)^{\frac{2024}{2025}}$

This post has been edited 1 time. Last edited by Tony_stark0094, Apr 12, 2025, 12:37 PM
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#9 h Apr 12, 2025, 12:29 PM • 1 Y

Y by khan.academy

Official Solution: We prove the result by induction with base case being $n = 2$ . Assume the problem holds for $n = k$ . We prove it for $k+1$ .

Notice that $a_{k+1} > a_k$ for all $k$ . Now,

$a_{k+1}^{{2025}/{2024}} = a_{k+1} a_{k+1}^{{1}/{2024}} > a_{k+1} a_k^{1/2024} = a_{k}^{{2025}/{2024}} + 1 > k + 1.$
This completes the induction.

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#10 h Apr 12, 2025, 12:43 PM

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Tony_stark0094 wrote:

Thapakazi wrote:

@above

ThatApollo777 wrote:

$a_k + \frac{1}{\sqrt[2024]{a_k}} > k^{2024/2025} + \frac{1}{\sqrt[2025]{k}}$

How?

$a_k+\frac{1}{\sqrt[2024]{a_k}}$ is a increasing function and $a_k > k^{\frac{2024}{2025}}$

I understand why $a_k > k^{2024/2025}$ , but I don't know why increasing helps here. Could you please elaborate more? Thanks!

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#11 h Apr 12, 2025, 12:50 PM

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Quote:

I understand why $a_k > k^{2024/2025}$ , but I don't know why increasing helps here. Could you please elaborate more? Thanks!

$f$ is increasing so by defination of increasing functions we can say that $a>b \implies f(a) >f(b)$ using $a_k > k^{2024/2025}$ we get
$\implies a_k+\frac{1}{\sqrt[2024]{a_k}} >k^{\frac{2024}{2025}}+\frac{1}{\sqrt[2025]{k}}=\frac{k+1}{\sqrt[2025]{k}}$

This post has been edited 1 time. Last edited by Tony_stark0094, Apr 12, 2025, 12:50 PM
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#12 h Apr 12, 2025, 12:54 PM

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Tony_stark0094 wrote:

Quote:

I understand why $a_k > k^{2024/2025}$ , but I don't know why increasing helps here. Could you please elaborate more? Thanks!

$f$ is increasing so by defination of increasing functions we can say that $a>b \implies f(a) >f(b)$ using $a_k > k^{2024/2025}$ we get
$\implies a_k+\frac{1}{\sqrt[2024]{a_k}} >k^{\frac{2024}{2025}}+\frac{1}{\sqrt[2025]{k}}=\frac{k+1}{\sqrt[2025]{k}}$

Ah clever. Thanks!

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cursed_tangent1434

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#13 h Apr 16, 2025, 5:43 AM • 1 Y

Y by khan.academy

Different inductive solution here. The base case $n=2$ is trivial as $a_2=2$ and $2^{2025}>2^{2024}$ . We assume the result holds for some $n \ge 2$ and show it also does for $n+1$ . Note,

$\begin{align*} a_{n+1}^{2025} &= \left(a_n + a_n^{-\frac{1}{2024}}\right)^{2025}\\ &= a_n^{2025} + \binom{2025}{1}a_n^{2024}\cdot a_n^{-\frac{1}{2024}} + \dots + \binom{2025}{2024}a_n\cdot a_n^{-\frac{2024}{2024}}+ a_n^{-\frac{2025}{2024}}\\ &= a_n^{2025} + \left[\sum_{i=1}^{2024} \binom{2025}{i} a_n^{2025-i}\cdot a_n^{-\frac{i}{2024}} \right]+ a_n^{-\frac{2025}{2024}} \end{align*}$
We now show a key inequality.

Claim : For all positive integers $1 \le i \le 2024$ we have the inequality,
$\binom{2025}{i} a_n^{2025-i}\cdot a_n^{-\frac{i}{2024}} \ge \binom{2024}{i}n^{2024-i}$

Proof : The desired inequality is equivalent to,
$\begin{align*} \binom{2025}{i} a_n^{2025-i}\cdot a_n^{-\frac{i}{2024}} &\ge \binom{2024}{i}n^{2024-i} \\ \binom{2025}{i} a_n^{\frac{2025(2024-i)}{2024}} &\ge \binom{2024}{i}n^{2024-i}\\ \frac{2025}{2025-i} a_n^{\frac{2025(2024-i)}{2024}} & \ge n^{2024-i} \end{align*}$ However, this is clear as for all $1 \le i \le 2024$ ,
$\frac{2025}{2025-i} a_n^{\frac{2025(2024-i)}{2024}} \ge a_n^{\frac{2025(2024-i)}{2024}} > n^{\frac{2024(2024-i)}{2024}} = n^{2024-i}$ as desired.

Thus,
$\begin{align*} a_{n+1}^{2025} & = a_n^{2025} + \left[\sum_{i=1}^{2024} \binom{2025}{i} a_n^{2025-i}\cdot a_n^{-\frac{i}{2024}} \right]+ a_n^{-\frac{2025}{2024}} \\ & > n^{2024} + \sum_{i=1}^{2024} \binom{2024}{i}n^{2024-i} + a_n^{-\frac{2025}{2024}} \\ & > (n+1)^{2024} \end{align*}$ which completes the induction.

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