$n$ 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.
 A: Since I am receiving a lot of flack for my poor use of inequalities, I will try to improve my method. First, note that you are analyzing the Harmonic Series $H_n$ and:
$$H_n \approx ln(n)+\gamma$$
However,
$$H_n < ln(n) +\gamma$$ 
For all $n$. Luckily,
$$H_n \approx ln(n) +\gamma - \frac{1}{n}$$
and
$$H_n > ln(n) +\gamma - \frac{1}{n}$$
for all $n$, so we have
$$ln(n) +\gamma + \frac{1}{n}<H_n<ln(n) +\gamma$$
If you want to see just how good these approximations are, look at them all graphed: https://www.desmos.com/calculator/a5ibmgvsdy
We would like to show for which the smalles value of $n$, $H_n \leq 4$ so lets solve both sides of the equality for $4$.
$$4<ln(n) + \gamma$$
$$4- \gamma < ln(n)$$
$$e^{4-\gamma}<n$$
$$30.6546491214 \lessapprox n$$ 
Now for the other side
$$ln(n) +\gamma - \frac{1}{n}<4$$
$$ln(n)-\frac{1}{n}<4-\gamma$$
And after analyzing the graph of $ln(n) + \frac{1}{n}$ (since $n$ isn't separable)
$$n \lessapprox 31.6$$
So finally we have
$$30.6546491214\lessapprox n \lessapprox 31.6$$
Therefore if $n$ is an integer, we can approximate
$$H_{31} \geq 4$$
I put this into Desmos for you as well so you can verify it is true: https://www.desmos.com/calculator/ttjyfn6ph3
A: Not assuming as much knowledge about the Harmonic series as the other answers, we can also do an integral approximation by Riemann sum.
Since $f(x) = 1/x$ is monotonic decreasing
$$\sum_{k=1}^N 1/k \approx 1+\int_{1}^{N}{\frac{1}x dx} = 1+\left[\log(x)\right]_1^N = \log(N)-\log(1)+1 = 4$$
But because of systematic underestimation due to the monotonic nature of our log function we might as well choose one displaced by one step and get systematic overestimation. The mean value of these will be $$(\exp(3)+\exp(3+1))/2 \approx 37$$
But the nature of this mean involving exponential functions, a geometric mean might be more suitable than an arithmetic mean:
$$\sqrt[2]{e^3 \cdot e^4} = e^{3.5}\approx 33$$
The real value appears to be $31$ or $32$.
A: 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 $27.83$.
A: There are good approximations as shown by the previous answers. Doing it intelligently you get very good estimates. However, it seems to me that the exact value of $n$ can only be determined by explicit calculations. I used Excel:
 $\quad$ 
Here is the complete table up to $n = 32$. Note that Excel uses $8$ decimals, thus the rounding error in these calculations does not affect our result $n = 31$.

