The Reciprocal Of Infinity An Infinite Number Of Times

Expression can be written as: $\lim_{n\to \infty} \frac 1n \displaystyle \sum_{r=1}^{\infty} \left(\dfrac{1}{1+\frac rn}\right)$ $~~~~~~~~~~~~~\Large =\int_{0}^{1}\dfrac{dx}{1+x}~~$ ( See this ) $=\huge \ln 2$

$a_n =\sum_{i=1}^{2N} \frac{1}{n} - \sum_{i=1}^N \frac{1}{n} = H_{2n} - H_{n}$

Where $H_n$ is a Harmonic number. Now from the integral test we have:

$\int_{n}^{n+1}\frac{dt}{t} \leq \frac{1}{n} \leq \int_{n-1}^{n} \frac{dt}{t}$

$\int_{1}^{N+1}\frac{dt}{t} \leq \sum_1^N \frac{1}{n} \leq 1 + \int_{1}^{N} \frac{dt}{t}$

$\ln(N+1) - \ln{1} \leq H_n \leq 1 + \ln(N) - \ln{1}$

$\ln(N+1) \leq H_n \leq 1 + \ln(N)$

$\ln(2N+1) \leq H_{2n} \leq 1 + \ln(2N)$

Taking the difference we get:

$\ln(2N+1) - \ln(N+1) \leq H_{2n} - H_{n} \leq 1 + \ln(2N) - \left(1 + \ln(N) \right)$ $\ln(\frac{2N+1}{N+1}) \leq H_{2n} - H_{n} \leq \ln(2)$

$\lim_{n \to \infty} \frac{2n+1}{n+1} = 2$ (L'Hopital)

Therefore by the Squeeze Theorem we get:

$\ln(2) \leq a_n \leq \ln(2)$

$\because lim_{n \to \infty} a_n = \ln{2}$

In general: For two Harmonic numbers $H_{pn}$ and $H_{qn}$ where $p \ge q$

$\lim_{n \to \infty} H_{pn} - H_{qn} = ln(\frac{p}{q})$

First calculate $\sum_{n=1}^{10} \Large \frac {1}{10+n}$ = $0.69315$
$\sum_{n=1}^{100} \Large \frac {1}{100+n}$ = $0.69315$
$\sum_{n=1}^{1000} \Large \frac {1}{1000+n}$ = $0.69315$
$\sum_{n=1}^{10000} \Large \frac {1}{10000+n}$ = $0.69315$
$ln(2) = 0.69315$ .
So, then answer is $ln(2)$ . $_\square$

Riemann sum form of the integral of $\frac{dx}{x}$ from 1 to 2