Hugely Harmonic

Relevant wiki: Harmonic Number

$\begin{aligned} \sum_{k=1}^N \frac 1k & > 100 \\ \implies \color{#3D99F6} H_N & > 100 & \small \color{#3D99F6} \text{where }H_n \text{ is the }n\text{th harmonic number.} \\ \color{#3D99F6} \gamma + \ln N & > 100 & \small \color{#3D99F6} \text{Approximate from }\gamma = \lim_{n \to \infty} (H_n - \ln n) \\ \ln N & > 100 - \gamma & \small \color{#3D99F6} \text{where }\gamma \text{ is Euler-Mascheroni constant.} \\ \log_{10} N & > (100 - \gamma)\log_{10} e \\ & > (100 - 0.5772)\log_{10} e \\ & > 43.179 \end{aligned}$

The number of digits of $N$ , $n = \left \lfloor \log_{10} N \right \rfloor + 1 = 43+1 = \boxed{44}$ .

For any decreasing function $f \$ :

$\displaystyle \sum_{k \ = \ a}^{N} f(k) \ \geq \ \int_{x \ = \ a}^{N + 1} f(x) \ dx$

For $f(x) = \dfrac 1x$ , we have

$\displaystyle \sum_{k \ = \ 1}^{N} \frac 1k \ \geq \ \int_{1}^{N + 1} \frac 1x \ dx \\ \sum_{k \ = \ 1}^{N} \frac 1k \ \geq \ \ln \left(N + 1\right)$

Because the growth of the harmonic series is so slow, we can say that $\displaystyle \sum_{k \ = \ 1}^{N + 1} \frac 1k \ \cancel{\geq} \ 101$ . Thus,

$\displaystyle \Bigg\lfloor{\sum_{k \ = \ 1}^{N} \frac 1k\Bigg\rfloor} = 100 \\ \implies 100 \ \leq \ \sum_{k \ = \ 1}^{N} \frac 1k \ < \ 101$

For $\displaystyle \sum_{k \ = \ 1}^{N} \frac 1k = 100$ , we have

$\displaystyle \begin{aligned} \ln\left(N + 1\right) & > 100 \\ \implies N & > e^{100} - 1 \ \approx \ 2.7 \times 10^{43} \end{aligned}$

In the extreme case that $\displaystyle \sum_{k \ = \ 1}^{N} \frac 1k = 101$ , we have

$\displaystyle \begin{aligned} \ln\left(N + 1\right) & < 101 \\ \implies N & < e^{101} - 1 \ \approx \ 7.3 \times 10^{43}\end{aligned}$

Thus, $2.7 \times 10^{43} < N < 7.3 \times 10^{43}$ , the numbers of which all have $43 + 1 = \boxed{44}$ digits.

The exact number is given here
as 15092688622113788323693563264538101449859497
which is 44 digits in length

We can rexpress the sum as:

$\sum_{x=0}^n\frac{1}{x}\approx\int_{1}^{N}\frac{1}{x}dx$

Evalutating the integral:

$\int_{1}^{N}\frac{1}{x}dx =\ln{|x|} \big|_1^N$

$\ln{|x|} \big|_1^N=\ln{N}-\ln{1}$

$\ln{N}-\ln{1}=\ln{N} - 0 = \ln{N}$

We know that this value must be greater than 100, so

$\ln{N} > 100$

Consequently:

$N=e^{100}$

The next step requires a bit of intuition:

We're told that $\log_{10}{e}\approx.434$

Furthermore, we know that 1 digit requires $10^1$ power. The number of power of 10's in a number is the $\log_10$ of that number.

We can think of this like e having .434 digits. Hence $e^{100}$ would have $\log_{10}{e^{100}}$ digits.

$\log_{10}{e^{100}}=100\log_{10}{e}=100*.434=43.4$

Because numbers can only have a whole number of digits, and N has at least 43.4 digits, it must have 44 digits.