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? 

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 $n = 12367$.
