Finding the smallest n satisfying $S_n > 10$ Let $S_n = 1 + \frac 12 + \frac 13 + \cdots + \frac 1n$, where $n \in \{ 1,2,3,\cdots\}$ Find the smallest $n$ satisfying $S_n > 10$. 
Sorry, it's my first time asking and I don't know how to format this thing. I still don't see anything even after staring at this for really long. Any clues? 
 A: We have an asympotic expansion for $H_n$:
$$H_n = \ln n + \gamma + \frac{1}{2n} - \sum_{k=1}^{\infty} \frac{B_{2k}}{2 k n^{2k}} = \ln n + \gamma + \frac{1}{2n} -\frac{1}{12 n^2}+ \frac{1}{120 n^4} - \dots$$
Reference : Wikipedia on Harmonic number
So
$$\ln n + \gamma + \frac{1}{2n} -\frac{1}{12 n^2} < H_n < \ln n + \gamma + \frac{1}{2n}$$
Solving $\ln n + \gamma = 10$ for $n$, we find $n \approx 12367$. With this value of $n$, we compute 
$$ H_n > \ln n + \gamma + \frac{1}{2n} -\frac{1}{12 n^2} = 10.000043$$
and $$H_{n-1} < \ln (n-1) + \gamma + \frac{1}{2(n-1)} = 9.999962144$$
so  $H_{n-1} < 10$ and $H_n > 10$ for $n = 12367$.
A: Should be near 
$$ \lfloor{ \frac{e^9}{2} }+1 $$
I use 
$$ \int_k^{k+1} \frac{1}{t} dt \leq \frac{1}{k} \leq \int_{k-1}^{k} \frac{1}{t} dt $$
You sum and integrate which give you log then you solve for upper and lower bound and verify.
A: $$12367$$
See my graph here as a show of that.
By calculator:$$S_{12367}=10.00004301$$
A: The $n$-th harmonic number $H_n=1+\frac{1}{2}+\dots+\frac{1}{n}$ is asymptotic to $\ln n+\gamma$, where $\gamma$ is the Euler-Mascheroni constant, equal to $0.57721\dots$
So, if you are searching for an $n$ such that $H_n>1$, you should solve the following equation:
$$
\ln n + \gamma > 10,
$$
and the first integer greater than $e^{10-\gamma}$ is $n=12367$.
This is an estimate of the exact value, one needs some error estimates to find it.
