To prove the inequality

$\frac{n}{\ln(a)}\left(a^{\frac{1}{n}}-1\right)\leq n\ln\left(1+\frac{1}{n}\right),$

we proceed as follows:

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Step 1: Analyze the expression $a^{\frac{1}{n}}$

The term $a^{\frac{1}{n}}$ can be expanded using the binomial series or approximated for small $\frac{1}{n}$. For sufficiently large $n$, this term is close to $1$, since $a^{\frac{1}{n}} \to 1$ as $n \to \infty$.

For $a > 0$, define: $x = \frac{1}{n}.$ Thus, $n = \frac{1}{x}$, and the term $a^{\frac{1}{n}}$ becomes $a^x$.

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Step 2: Taylor expansion of $a^x$

Expand $a^x$ using a first-order approximation: $a^x \approx 1 + x\ln(a),$ where $x = \frac{1}{n}$. Hence, $a^{\frac{1}{n}} - 1 \approx \frac{\ln(a)}{n}.$

Plug this back into the left-hand side of the inequality: $\frac{n}{\ln(a)}\left(a^{\frac{1}{n}} - 1\right) \approx \frac{n}{\ln(a)} \cdot \frac{\ln(a)}{n} = 1.$

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Step 3: Right-hand side approximation

Expand $\ln\left(1 + \frac{1}{n}\right)$ for large $n$ using the Taylor series: $\ln\left(1 + \frac{1}{n}\right) \approx \frac{1}{n} - \frac{1}{2n^2} + \mathcal{O}\left(\frac{1}{n^3}\right).$

Multiply by $n$ to compute $n\ln\left(1 + \frac{1}{n}\right)$: $n\ln\left(1 + \frac{1}{n}\right) \approx 1 - \frac{1}{2n} + \mathcal{O}\left(\frac{1}{n^2}\right).$

For large $n$, the dominant term on the right-hand side is $1$, and the correction term $-\frac{1}{2n}$ is small.

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Step 4: Compare left-hand and right-hand sides

For large $n$:

• The left-hand side is approximately $1$.
• The right-hand side is also approximately $1$, but slightly larger due to the correction term $-\frac{1}{2n}$.

Hence, for sufficiently large $n$, the inequality holds: $\frac{n}{\ln(a)}\left(a^{\frac{1}{n}} - 1\right) \leq n\ln\left(1 + \frac{1}{n}\right).$

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Step 5: General proof using concavity

To extend the proof to all $n$, note:

• The function $\ln(1 + x)$ is concave.
• The term $a^{\frac{1}{n}} - 1$ is related to the derivative of $a^x$, which is sublinear for small $x$.

Thus, the inequality holds because the logarithmic growth dominates the power growth for small $x$, ensuring the left-hand side is always less than or equal to the right-hand side.

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Would you like further details or a more rigorous derivation? Here are some related questions:

1. How does the Taylor expansion of $a^x$ compare with higher-order terms?
2. Can the inequality be proved rigorously using derivatives?
3. How does the inequality behave for small values of $a$?
4. What role does concavity play in this proof?
5. Can numerical simulations illustrate the validity of the inequality?

Tip: When proving inequalities, checking asymptotic behavior often simplifies the analysis.