Is the following inequality correct and how to prove it? We have $0 < m_n < 1$ and $m_{n+1}=m_{n}(1-\dfrac{m_{n}}{n})$ for all $n \in \mathbb{N}, n \ge 2$.
Does the inequality below hold true for all $n \in \mathbb{N}, n \ge 2$, and if it does, how to prove it ?
$$
\sum_{k=1}^{n} \frac{1}{k}<\frac{1}{m_{n}}<\sum_{k=1}^{n} \frac{1}{k}+1
$$
 A: Update: I wonder where OP got this problem. I found this one: Probabilistic Sieve of Eratosthenes and they are related because the harmonic series can be approximated by the logarithm function as shown in that problem.

This is not a complete answer but it's too long to be replied in the comment section.
Results so far: LHS inequality is true (with proof) as long as $m_2 < \frac{2}{3}$. RHS inequality seems to be true if $m_2 \geqslant 0.61 (0.6096)$ and false if $m_2 \leqslant 0.6095$ but is not proved. (We also need $m_1 > \frac{1}{2}$ but its value is mostly irrelevant.)
$\forall n>1$, we have
$$
\frac{1}{m_{n+1}}-\frac{1}{m_n}=\frac{1}{m_n\left(1-\frac{m_n}{n}\right)}-\frac{1}{m_n} = \frac{1}{m_n} \left( \frac{1}{1-\frac{m_n}{n}} - 1\right)=\frac{\frac 1n}{1-\frac{m_n}{n}}> \frac 1n,
$$
$$
\frac{1}{m_{n+1}}-\frac{1}{m_n}=\frac{\frac 1n}{1-\frac{m_n}{n}} < \frac{\frac 1n}{1-\frac 1n}= \frac{1}{n-1} 
$$
Denote $S_n=\sum_{k=1}^n \frac 1k$, then $\forall N>2$ $$
S_{N-1}-1 < \frac{1}{m_N}-\frac{1}{m_2}  < \sum_{n=2}^{N-1} \frac{1}{n-1} = \sum_{n=1}^{N-2} \frac 1n = S_{N-2}. \tag 1
$$
The LHS inequality becomes
$$
\frac{1}{m_N} > \frac{1}{m_2} + S_N - 1 - \frac{1}{N} > \frac 12 -\frac{1}{N} + S_N \geqslant S_N, \text{ since } m_2<\frac 23.
$$
So long as $1+\frac 12 < \frac{1}{m_2}$, or $m_2<\frac{2}{3}$, then $S_N < \frac{1}{m_N}$ is always true.
The RHS inequality of (1) however, cannot be used to prove $\frac{1}{m_{n}}<S_n+1$.
This is because if $\frac{1}{m_2} + S_{N-2} < S_{N}+1$ then $\frac{1}{m_2} < 1 + \frac{1}{N-1} + \frac 1N \Rightarrow \frac{1}{m_2} \leqslant 1$, as $N \to \infty$, a contradiction.
Mathematical induction utilizing both sides also didn't work.
Suppose $S_N < \frac{1}{m_N} < S_N + 1$, then
$$
\frac{1}{m_{N+1}}=\frac{1}{m_N (1-\frac{m_N}{N})} <  \frac{S_N+1}{1-\frac{1}{NS_N}}
$$
Unfortunately $\frac{S_N+1}{1-\frac{1}{NS_N}}<S_{N+1}+1=S_N+\frac{1}{N+1}+1$ is false if you expand all terms and simplify.
Numerical results I did run a script in R and compared $1/m_N$ with $S_N+1$ up to $N=10^8$ for various values of $m_2$ and I found that $1/m_n < S_n+1$ for $n=2,3,\ldots, 10^8$ if $m_2>0.61$.
Here's the script:
N=10^8;
L = cumsum(1/(1:N)); # S_N
R = L+1; # S_N+1
m=rep(NA,N);
m[1]=0.51;
m[2] = 0.61; # only need to change this line
for(i in 3:N) m[i]=m[i-1]*(1-m[i-1]/(i-1));
r=1/m;
all(L<r & r<R) # this should return true
