Harmonic number inequality With the $n$th harmonic number defined as $$
H_n  = \sum\limits_{k = 1}^n  {\frac{1}{k}} 
$$ I'm supposed to find the minimum EXACT value of $n$ such that $H_n>100$ . I could only find approximations, but my professor demands an exact value . Is this possible ?
If so, how can we find it?
 A: We have:
$$ \log(n+1/2)+\frac{1}{24n^2}-\frac{1}{24n^3}\leq H_n-\gamma \leq \log(n+1/2)+\frac{1}{24n^2}$$
hence the smallest natural number for which $H_n>100$ is about $m=\left\lfloor e^{100-\gamma}-\frac{1}{2}\right\rfloor.$

Notice that $\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}=-\gamma+H_{x-1}$ is a concave function, since:
$$ \psi'(x) = \sum_{n\geq 0}\frac{1}{(n+x)^2}.$$
This gives that the Newton's method with starting point $m$ is a really wise choice to find the exact value of the least integer $n$ such that $H_n>100$.
A: The simplest pertinent estimate is based on Riemann sums. Specifically, if the step size is $h=1$, then $H_n$ is the corresponding upper Riemann sum for 
$$\int_1^n \frac{1}{x} dx.$$
Hence $H_n \geq \ln(n)-\ln(1) = \ln(n)$. Solving $\ln(n) \geq 100$ we have $n \geq e^{100}$. 
For the other bound, we have that
$$H_n - 1 \leq \int_1^n \frac{1}{x} dx \Rightarrow H_n \leq \ln(n)+1.$$
Solving $\ln(n)+1 \leq 100$ we have $n \leq e^{99}$, so your minimal answer will be between $e^{99}$ and $e^{100}$. 
This is clearly quite crude. However, about the only way to improve this is to accurately estimate the Euler-Mascheroni constant $\gamma$, since when $n$ is large we have $H_n \approx \ln(n) + \gamma$, meaning that your minimal answer should be in the vicinity of $e^{100-\gamma}$. 
A: The answer is $$\lceil\chi\rceil$$
where $\chi$ is the solution of $$\text{N}(\chi)=100$$
where $\text{N}$ is the analytic continuation of the harmonic sum function, and $\lceil\rceil$ is the integer ceiling function.
This can be quickly calculated in Mathematica:
Ceiling[x] /. Solve[HarmonicNumber[x] == 100, x][[1]]

which instantly yields the answer:
15092688622113788323693563264538101449859497

