solving natural log inequality

Question

How can I show that

$0 \le (\sum_{x=1}^{n}\frac{1}{x})-ln(n) \le 1 - \frac{1}{n}$

Do I raise both sides by $e$ or perhaps take integral of both sides? If so, I'm still not quite sure where to go from there

Answer 1

Start with a basic integration of $\frac{1}{x}$ from $n-1$ to $n$.

$$\int_{n}^{n+1}\frac{1}{n} - \frac{1}{x}dx$$

Since $\frac1n$ is the lower limit of this decreasing function, the value of the integrand will always be greater than or equal to $0$ through out the interval of integration.

Now if you sum this up from $n = 1$ to $n-1$, you get

$$\int_{1}^{n}\left(\frac{1}{1} + \frac{1}{2}+ \frac{1}{3}...\frac{1}{n-1}\right) - \frac{1}{x} dx$$

$$\int_{1}^{n} \left(\sum_{i=1}^{n-1}\frac{1}{i}\right)- \frac{1}{x}dx$$

$$\left(\sum_{i=1}^{n-1}\frac{1}{i}\right) - \log n$$

Sum of non negative quantities always give a non negative quantity, and hence the above quantity is $\geq 0$, thus proving the first part of the inequality.

For the second part, sketch out the box using the graph of $\frac1x$ from $n$ to $n+1$ on x-axis with $\frac1n$ to $\frac{1}{n+1}$ on y-axis. Now if you shade the area of this box that pertains to the integrand in the first equation, you'll notice that it is always less than equal to the area of the box.

$$\left(\int_{n}^{n+1}\frac{1}{n} - \frac{1}{x}dx\right) \leq A_{box}$$

$$A_{box} = \left(\frac{1}{n} - \frac1{n+1}\right)(n+1 - n) = \left(\frac{1}{n} - \frac1{n+1}\right)$$

Similarly as before, sum the inequality on both sides from $1$ to $n-1$ to get

$$\left(\sum_{i=1}^{n-1}\frac{1}{i}\right) - \log n \leq (1 -\frac12) + (\frac12 - \frac13) + (\frac13 - \frac14) ... (\frac1{n-1} -\frac1n) $$

$$\left(\sum_{i=1}^{n-1}\frac{1}{i}\right) - \log n \leq 1 - \frac1n $$

Thus, proving the second part of the inequality

Answer 2

Actually we have $$\displaystyle\sum_{k=1}^{n}\frac{1}{k+1}<\ln (n+1)<\displaystyle\sum_{k=1}^{n}\frac{1}{k}$$ which maybe is what you exactly want.

Notice that we only need to prove $$\frac{1}{k+1}<\ln (1+\frac{1}{k})<\frac{1}{k}\quad\forall k\in \mathbb{Z}^+$$

which can be yielded by $$(\frac{k+1}{k})^k<e<(\frac{k+1}{k})^{k+1}$$

The inequality can be revealed by monotonicity of $(1+\frac{1}{n})^{n}$ and $(1+\frac{1}{n})^{n+1}$ .

And for this we have an interesting elementary proof as follows,

we have $$\ln \left(1+\tfrac{1}{n+1}\right) = \int_{1}^{1+\frac{1}{n+1}}\frac{1}{x}\,dx>\frac{1}{n+2}=\frac{n\left(n+1\right)}{n+2}\int_{1+\frac{1}{n+1}}^{1+\frac{1}{n}}\,dx > n\int_{1+\frac{1}{n+1}}^{1+\frac{1}{n}}\frac{1}{x}\,dx$$ Now add $$n\ln \left(1+\tfrac{1}{n+1}\right) = n\int_{1}^{1+\frac{1}{n+1}}\frac{1}{x}\,dx$$ We get $$\left(n+1\right)\ln \left(1+\tfrac{1}{n+1}\right) > n\int_{1}^{1+\frac{1}{n}}\frac{1}{x}\,dx = n\ln \left(1+\tfrac{1}{n}\right)$$ which means $(1+\frac{1}{n})^{n}$ is strictly increasing.
The monotonicity of the other sequence can be similarly proved.

Added:(another elementary proof of the monotonicity)

$$ n\ln \left(1+\tfrac{1}{n}\right)=\int_{0}^{1}\frac{n}{x+n}\,dx=\int_{0}^{1}\frac{1}{\frac{x}{n}+1}\,dx$$

hence it is clear that the RHS is increasing, since for any $x\in(0,1)$ and any $N>n$ we have $$ \frac{1}{\frac{x}{n}+1}<\frac{1}{\frac{x}{N}+1}$$ so we have $(1+\frac{1}{n})^{n}$ is strictly increasing.
The monotonicity of the other sequence can be similarly proved.