Question

Show that for each integer \(\mathrm{m}>1\), \(\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{m}<\ln m<1+\frac{1}{2}+\ldots+\frac{1}{m-1}\)

Short answer

Yes, the inequality holds true for all integers m > 1, as proven using mathematical induction and the Taylor series for the natural logarithm function.

Step-by-step solution

Introduction

The given exercise is to prove the following inequality: $$\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{m}<\ln m<1+\frac{1}{2}+\ldots+\frac{1}{m-1}$$ for all integers \(m > 1\). We will use induction to prove this inequality.

Base case

Let's check the base case when \(m = 2\): $$\frac{1}{2}<\ln 2<1$$ Since \(\frac{1}{2} = 0.5\) and \(\ln2 \approx 0.6931\), the inequality holds for \(m = 2\). Thus, the base case is valid.

Inductive step

Now, let's assume that the inequality holds for some arbitrary integer \(k > 1\): $$\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{k}<\ln k<1+\frac{1}{2}+\ldots+\frac{1}{k-1}$$ We want to show that it also holds for \(k + 1\): $$\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{k}+\frac{1}{k+1}<\ln (k+1)<1+\frac{1}{2}+\ldots+\frac{1}{k}$$

Using the inductive hypothesis

Let's consider the inequality from the inductive step: $$\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{k}+\frac{1}{k+1}<\ln (k+1)$$ According to the inductive hypothesis, we can replace the left-side of the inequality with \(\ln k\): $$\ln k+\frac{1}{k+1}<\ln (k+1)$$ To show this, we can use the property of logarithm: \(ln(k + 1) - ln(k) = ln(\frac{k+1}{k})\). Thus, we have to show: $$\frac{1}{k+1}<\ln \left(\frac{k+1}{k}\right)$$

Using the Taylor series for the natural logarithm

We can use the Taylor series expansion of \(\ln(1 + x)\) around \(x = 0\) to prove our inequality: $$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$$ Since \(x=\frac{1}{k}\) and \(k>1\), we have: $$\ln \left(\frac{k+1}{k}\right) = \frac{1}{k} - \frac{1}{2k^2} + \frac{1}{3k^3} - \cdots$$ Note that each term in the series is positive and smaller than \(\frac{1}{k+1}\). Therefore, we get: $$\frac{1}{k+1}<\ln \left(\frac{k+1}{k}\right)$$

Checking the other side of the inequality

According to the inductive hypothesis, inequality holds for \(k\): $$1 + \frac{1}{2} +\ldots+\frac{1}{k-1}<\ln k$$ Now, we want to show that this inequality holds for \(k+1\): $$1+\frac{1}{2}+\ldots+\frac{1}{k}<\ln (k+1)$$ Since we already proved that \(\frac{1}{k+1}<\ln(\frac{k+1}{k})\), we can write: $$1+\frac{1}{2}+\ldots+\frac{1}{k}< \ln k + \frac{1}{k+1} \leq \ln k + \ln \left(\frac{k+1}{k}\right) = \ln (k+1)$$

Conclusion

Now, we have proved that the inequality holds for the integer \(k+1\): $$\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{k+1}<\ln (k+1)<1+\frac{1}{2}+\ldots+\frac{1}{k}$$ Thus, by mathematical induction, we have shown that the inequality holds for all integers \(m > 1\).