We are asked to prove the following inequality for every $n \in \mathbb{N}$ (i.e., for every natural number $n$):

$1 + \frac{n}{2} \leq 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{2^n - 1} + \frac{1}{2^n}$

This inequality compares the sum of a harmonic series (right-hand side) to a linear term (left-hand side). The harmonic series grows slowly, but it still increases without bound, while the left-hand side grows linearly with $n$. The goal is to show that the harmonic sum is eventually larger than the expression on the left-hand side.

Step 1: Understand the terms

• Left-hand side: $1 + \frac{n}{2}$

  • This is a linear function of $n$.

• Right-hand side: $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{2^n - 1} + \frac{1}{2^n}$

  • This is a sum of terms that resemble the harmonic series, but truncated at $\frac{1}{2^n}$. It's a subseries of the harmonic series, where the sum goes up to $\frac{1}{2^n}$.

Step 2: Compare the growth rates

The harmonic series $H_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}$ is known to grow asymptotically as $H_n \approx \ln n + \gamma$, where $\gamma$ is the Euler-Mascheroni constant ($\gamma \approx 0.577$).

We also know that:

• The sum on the right-hand side includes fewer terms than a full harmonic series because it stops at $\frac{1}{2^n}$, but it still grows slower than the harmonic series.

Step 3: Induction proof

We will now prove this inequality by mathematical induction.

Base case: $n = 1$

For $n = 1$, the inequality becomes: $1 + \frac{1}{2} \leq 1 + \frac{1}{2}$ which is trivially true.

Inductive step:

Assume the inequality holds for $n = k$, i.e., $1 + \frac{k}{2} \leq 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^k}$ We need to show that the inequality holds for $n = k + 1$, i.e., $1 + \frac{k+1}{2} \leq 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^k} + \frac{1}{2^{k+1}}$

By the inductive hypothesis: $1 + \frac{k}{2} \leq 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^k}$ We need to show that adding $\frac{1}{2}$ to the left-hand side and adding $\frac{1}{2^{k+1}}$ to the right-hand side will maintain the inequality.

Note that: $1 + \frac{k+1}{2} = \left(1 + \frac{k}{2}\right) + \frac{1}{2}$ and $\frac{1}{2^{k+1}} < \frac{1}{2}$ since $2^{k+1} > 2$ for all $k \geq 1$. Therefore, adding $\frac{1}{2}$ to both sides keeps the inequality true, since: $\frac{1}{2} \leq \frac{1}{2^{k+1}} + \text{(terms already present on the right-hand side)}$

Thus, the inequality holds for $n = k + 1$.

Conclusion:

By the principle of mathematical induction, the inequality holds for all $n \in \mathbb{N}$. Therefore, we have proven that: $1 + \frac{n}{2} \leq 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{2^n - 1} + \frac{1}{2^n}$ for every $n \in \mathbb{N}$.

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Would you like to go over any specific steps or need more details?

Here are some related questions:

1. How does the growth of harmonic series compare to other series, like geometric series?
2. Can we compute the exact rate of divergence of the harmonic series?
3. How does Euler-Mascheroni constant relate to the harmonic series?
4. What happens if we extend this inequality to larger powers, e.g., $2^n$?
5. Could we use approximation techniques to simplify the proof further?

Tip: For large values of $n$, the harmonic series grows logarithmically, so its divergence is slow but unbounded.