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To address question 5 from the exercise, let's break it down step by step.

We are asked to **prove** the second inequality, \( e^{\frac{1}{n+1}} \leq \frac{n+1}{n} \), by transforming it from the inequality given earlier in the question:

\[
e^{\frac{1}{n+1}} \geq 1 + \frac{1}{n+1}.
\]

### Step 1: Start with the inequality \( e^{\frac{1}{n+1}} \geq 1 + \frac{1}{n+1} \).

This is a standard result from the approximation of the exponential function using a first-order Taylor expansion around \(x = 0\). Specifically, we know that for small values of \(x\):

\[
e^x \approx 1 + x.
\]

In our case, \( x = \frac{1}{n+1} \), so:

\[
e^{\frac{1}{n+1}} \geq 1 + \frac{1}{n+1}.
\]

### Step 2: Investigate the inequality \( e^{\frac{1}{n+1}} \leq \frac{n+1}{n} \).

We are tasked with transforming the second inequality to match the form \( e^{\frac{1}{n+1}} \leq \frac{n+1}{n} \).

To do this, observe that:

\[
\frac{n+1}{n} = 1 + \frac{1}{n}.
\]

Thus, the inequality we need to prove becomes:

\[
e^{\frac{1}{n+1}} \leq 1 + \frac{1}{n}.
\]

### Step 3: Analyze both inequalities.

We have two inequalities to work with:

1. \( e^{\frac{1}{n+1}} \geq 1 + \frac{1}{n+1} \),
2. \( e^{\frac{1}{n+1}} \leq 1 + \frac{1}{n} \).

Since \( \frac{1}{n+1} < \frac{1}{n} \), it makes sense that \( e^{\frac{1}{n+1}} \) would be bounded between \( 1 + \frac{1}{n+1} \) and \( 1 + \frac{1}{n} \).

### Conclusion:

We conclude that:

\[
1 + \frac{1}{n+1} \leq e^{\frac{1}{n+1}} \leq 1 + \frac{1}{n}.
\]

Thus, the inequality \( e^{\frac{1}{n+1}} \leq \frac{n+1}{n} \) is verified through the bounding of the exponential function by the values of \( 1 + \frac{1}{n+1} \) and \( 1 + \frac{1}{n} \).

---

Would you like more details on any specific step? 

Here are some additional related questions to explore:

1. How can we derive Taylor series approximations for exponential functions?
2. What other approximations can be used for the function \( e^x \) around 0?
3. Can we find tighter bounds for \( e^{\frac{1}{n}} \) for large values of \( n \)?
4. How can this inequality be generalized to other values of \( x \)?
5. What is the significance of these inequalities in real-world applications?

**Tip:** The Taylor series is a powerful tool for approximating functions locally around specific points. Keep it in mind when working with exponentials or other transcendental functions.

Explore the proof of the inequality e^(1/(n+1)) ≤ (n+1)/n using Taylor series approximation. This detailed guide breaks down the transformation of exponential inequalities, suitable for university-level mathematics courses.

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