How to prove that for each natural number ($N$) the following is true:
$$\ln(N+1)-\ln(N) \geq \frac{1}{N+1}$$
I tried using induction but it seems like the end of a road. The first case on ($N=0$) was easy but the rest not.
Note: I'm looking for solution using induction
On
$\ln (N+1) - \ln (N) =$
$\ln {\frac {N+1}N} = $
$\int_1^{\frac {N+1}N} \frac 1t dt$
Now $\frac {N+1}N> 1$ and for $1\le x \le \frac {N+1}N$ we have $1 \ge \frac 1x \ge \frac N{N+1}$
So $[\frac {N+1}N - 1]\cdot 1 \ge \int_1^{\frac {N+1}N} \frac 1t dt \ge [\frac {N+1}N - 1]\cdot \frac N{N+1}$
So $\frac 1N \ge \ln(N+1)-\ln N \ge \frac 1N\cdot \frac N{N+1}=\frac 1{N+1}$
On
This way is a bit long-winded but it does work. Note: I started writing this before the request for a proof by induction was made.
Consider $$\ln(N+1)-\ln(N)$$
$$=\ln\bigg(\frac{N+1}{N}\bigg)$$
$$=\ln\bigg(1 +\frac{1}{N}\bigg)$$
$$=\sum_{j=1}^{\infty} \frac{(-1)^{j+1}\Big(\frac{1}{N}\Big)^j}{j}$$
The Taylor series converges if $|\frac{1}{N}|< 1$, so this works for $N\ge 2$. However, the cases for $N = 0,1$ are easy to verify by calculation.
For $N \ge 2$,
$$\ln\bigg(1 +\frac{1}{N}\bigg) = \frac{1}{(1)N} - \frac{1}{(2)N^2} + \frac{1}{(3)N^3} -\frac{1}{(4)N^4} + \cdots$$
$$= \Bigg(\frac{1}{(1)N} - \frac{1}{(2)N^2}\Bigg) + \Bigg(\frac{1}{(3)N^3} -\frac{1}{(4)N^4}\Bigg) + \cdots$$
It should be clear that each bracketed term is positive; thus,
$$\ln\bigg(1 +\frac{1}{N}\bigg) > \frac{1}{N} - \frac{1}{2N^2}$$
Now consider
$$\Bigg(\frac{1}{N} - \frac{1}{2N^2}\Bigg) - \Bigg(\frac{1}{N+1}\Bigg)$$
$$ = \Bigg(\frac{2N-1}{2N^2}\Bigg) - \Bigg(\frac{1}{N+1}\Bigg)$$
$$ = \frac{N-1}{(2N^2)(N+1)} $$
$$ > 0 $$ since $N \ge 2$. Therefore,
$$\frac{1}{N} - \frac{1}{2N^2} > \frac{1}{N+1} $$
$$\implies \ln\bigg(1 +\frac{1}{N}\bigg) \ge \frac{1}{N+1}$$
The area under the curve $y=\dfrac{1}{x}$ between $x=N$ and $x=N+1$ is greater than the area of a rectangle of width $1$ and height $\dfrac{1}{N+1}$.
Draw a sketch to see this.