<s>[INST] $n$ be the smallest positive integer such that $1+ \frac{1}{2} + \frac{1}{3} + \frac {1}{4}+\cdots \geq 4$. $n$ be the smallest positive integer such that $1+ \frac{1}{2} + \frac{1}{3} +\frac14+\cdots \geq 4$.
My Attempt : I can show that $ \log(n+1) \leq 1+ \frac{1}{2} + \frac{1}{3} + \frac {1}{4} +\cdots+ \frac {1}{n}$. And by using this inequality  I can say that the smallest number will be less than $54$. But I can not find the smallest number.  
Can anyone please help me by giving some hints.
 [/INST] We have
$$x-1<\lfloor x\rfloor\le x$$ and
$$\frac1x\le\frac1{\lfloor x\rfloor}<\frac1{x-1}.$$
Now integrating between $2$ and $n+1$, and adding $1$,
$$1+\log\frac{n+1}2\le\sum_{k=1}^n\frac 1k<1+\log n.$$
The lower and upper bounds equal $4$ for $n=41.17$ and $n=20.09$ and the desired solution will lie in that range.
We can narrow that range by computing more terms explicitly
$$1+\frac12+\frac13+\frac14+\frac15+\log\frac{n+1}6\le\sum_{k=1}^n\frac 1k<1+\frac12+\frac13+\frac14+\frac15+\log\frac n5,$$gives $32.40$ and $27.83$.
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