Find least $n$ such that $\sum_{2\leq k \leq n} \frac{1}{k} \geq 5$ 
Let $n$ be the least positive integer such that $\sum_{2\leq k \leq n} \frac{1}{k} \geq 5$. Choose the correct option:
(a) $n \leq 32$
(b) $32<n\leq96$
(c) $96<n\leq729$
(d) $729<n$

I know the series $\sum \frac{1}{n}$ diverges. But I am finding it difficult on how to find the solution to the above problem.
Any help is appreciated. Thanks in Advance.
 A: HINT : If you try to solve this sort of problem, basically you need some inequalities. 
Here, there's a famous one:
$\frac{1}{n} + \log n < 1 + \frac 1 2 + \cdots + \frac 1 n < 1 + \log n $ for $n \geq 2$.
To prove this, you can use integration.
A: Classical bounding on the Harmonic series using series - integral comparison is 
$$(\ln n)  \ge \sum_{2\leq k \leq n} \frac{1}{k} \geq \ln(n+1) - 1$$
As $\ln 96 < 5$ and $\ln 730 > 6 $, the only possible option is $(c)$.
A: If you know the properties of harmonic numbers
$$S_n=\sum_{k=2}^n\frac{1}{k}=-1+H_n$$
Now, assuming that $n$ is large, using the asymptotics
$$S_n=\log \left({n}\right)+\gamma -1+O\left(\frac{1}{n}\right)$$ then, as an approximation
$$\log \left({n}\right)+\gamma -1 \geq 5 \implies n \geq e^{6-\gamma } \approx 226.5$$
A: One can have even more precise bracketings. 
We can use for example the following Euler-Maclaurin expansion (due to Ramanujan) with first terms :
$$\sum_{k=1}^n \dfrac{1}{k} =\dfrac12\ln(2p)+\gamma+\dfrac{1}{12p}-\dfrac{1}{120 p^2}+\dfrac{1}{630p^3}-\dfrac{1}{1680p^4}+\cdots \tag{1}$$
Where $p:=\dfrac{n(n+1)}{2}$ and $\gamma\approx 0.57721566490153286060...$ is the Euler-Mascheroni constant.
giving for $n=227$ for the result with 
(This the third formula in "Asymptotic expansions" here) 
In particular, (1) gives very precise bracketings like this one:
$$\dfrac12\ln(2p)+\gamma+\dfrac{1}{12p}-\dfrac{1}{120 p^2}<\sum_{k=1}^n \dfrac{1}{k}< \dfrac12(1/2)\ln(2p)+\gamma+\dfrac{1}{12p}\tag{2}$$
according to the fact that the error done in truncating an alternate series  whose general term is decreasing is less (in absolute value) than the first omitted term.
Here is the answer : $\color{red}{n=227}$ because (2) gives :
$$6.004366708345565 < \sum_{k=1}^{\color{red}{227}}\dfrac{1}{k} < 6.004366708358010$$
whereas :
$$5.999961422001953  < \sum_{k=1}^{226} \dfrac{1}{k} <   5.999961422014619$$
(please note the very narrow bracketting : 11 significant digits !)
Therfore, the good answer was answer c).
