<s>[INST] How to prove that $\ln (N+1) - \ln (N) \geq \frac{1}{N+1}$ 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
 [/INST] 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,
$$\frac{1}{N} - \frac{1}{2N^2} > \frac{1}{N+1} $$
$$\implies \ln\bigg(1 +\frac{1}{N}\bigg) \ge \frac{1}{N+1}$$
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