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Using the fact: ${x\over 1+x}<\ln (1+x)<x$, when $x>0$; deduce that: \begin{equation} \ln{2n+1\over n+1}<{1\over n+1}+{1\over n+2}+\cdots +{1\over 2n}<\ln 2 \end{equation}

This problem also asked to prove the fact I have given. I could prove it using MVT, but the deduction which should be easier is out of my hand.

Saikai Prime
  • 1,133

2 Answers2

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Make a change of variables $y = \frac{x}{1+x}$, we have: $x = \frac{y}{1-y}$. So, from $\frac{x}{1+x}<\ln(x+1)$, we have : $$y<\ln(\frac{1}{1-y}) = -\ln(1-y) \tag{1}$$

And because $\ln(1+x)<x\tag{2}$

From (1) and (2), we deduce that

$$\ln(1+x)<x<-\ln(1-x)$$

Replacing $x$ by $\frac{1}{n+k}$ for $k=1,...,n$ $$\ln(n+k+1)-\ln(n+k) < \frac{1}{n+k}< -(\ln(n+k-1)-\ln(n+k)) $$ Hence $$\sum_{i=1}^n(\ln(n+k+1)-\ln(n+k))<\sum_{i=1}^n \frac{1}{n+k}<\sum_{i=1}^n(\ln(n+k)-\ln(n+k-1))$$ or $$\ln(2n+1)-\ln(n+1)<\sum_{i=1}^n \frac{1}{n+k}<\ln(2n)-\ln(n)$$ or $$\ln \left(\frac{2n+1}{n+1} \right)<\sum_{i=1}^n \frac{1}{n+k}<\ln(2)$$

NN2
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The fact is: $\frac{x}{1+x} \lt \ln(1+x) \lt x$, when $x > 0$. Since we particularly know that $\ln(1+x) \lt x$, we can write: $$\begin{align} & \ln(2n+1)-\ln(2n)=\ln\Biggl(1+\frac{1}{2n}\Biggl)\lt\frac{1}{2n} \\ & \ln(2n)-\ln(2n-1)=\ln\Biggl(1+\frac{1}{2n-1}\Biggl) \lt \frac{1}{2n-1} \\ & \space \space \space \space \space \vdots \\ & \ln(n+2)-\ln(n+1)=\ln\Biggl(1+\frac{1}{n+1}\Biggl)\lt \frac{1}{n+1} \\ \end{align} $$ If we rewrite these expressions in a single line all at once, we eventually get: $$\ln\frac{2n+1}{n+1}\lt \frac{1}{n+1} + \frac{1}{n+2} + \cdots +\frac{1}{2n}$$ $\ln()$ is an increasing function where it is defined. We can write for any $x>\frac{1}{2}$ that: $$\ln\Biggl(2-\frac{1}{x}\Biggl)\lt \ln2 \lt \ln\Biggl(2+\frac{1}{x} \Biggl)$$ Therefore, we can deduce that: $$\ln\frac{2n+1}{n+1}=\ln\Biggl(2-\frac{1}{n+1}\Biggl) \lt \ln2 \lt \ln\Biggl(2+\frac{1}{n+1} \Biggl)$$ Finally: $$\ln\frac{2n+1}{n+1} \lt \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} \lt \ln2$$

Fatma Keskin
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