Finding the smallest n satisfying $S_n > 10$

Question

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?

Answer 1

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 $n = 12367$.

Answer 2

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.

Answer 3

$$12367$$

See my graph here as a show of that.

By calculator:$$S_{12367}=10.00004301$$

Answer 4

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.