Limits + Summation = Fun

We can use Riemann Sums (We can prove riemann integrability in the interval $[1,\frac{1}{N}]$ of $\displaystyle \left\{\frac{1}{x}\right\}$ because the set of points of discontinuity will have a finite derived set which will be the singleton set containing only $0$ ) to convert it into an integral :-

$\lim_{N\to\infty}\int_{\frac{1}{N}}^{1}\left\{\frac{1}{x}\right\}dx$ .

Now let us write $\displaystyle \left\{\frac{1}{x}\right\}= \frac{1}{x} - \left\lfloor \frac{1}{x}\right\rfloor$ and look at the integral:-

$\int_{\frac{1}{N}}^{1}\left(\frac{1}{x} - \left\lfloor \frac{1}{x}\right\rfloor\right)dx$

Now we can use the following property of the Greatest Integer function:-

$\dfrac{1}{n} < x < \dfrac{1}{n-1}\implies n-1<\dfrac{1}{x}<n \text{ for } n>1 \implies \left\lfloor \frac{1}{x} \right\rfloor = n-1$

So we arrive at the following :-

$\ln(N) - \sum_{r=1}^{N-1}r\left(\frac{1}{r}-\frac{1}{r+1}\right) = \ln(N) - \sum_{r=1}^{N-1}\frac{1}{r+1} = \ln(N)-H_{N} +1$ .

Now making $\displaystyle N\to\infty$ and using the fact that $\lim_{n\to\infty}\left( H_{n}-\ln(n)\right) = \gamma$ . We have our answer as :- $1-\gamma$ .

Notations:-

$\gamma$ denotes the Euler-Mascheroni Constant .

$H_{n}$ denotes the nth Harmonic Number

$\lfloor \cdot \rfloor$ denotes the Floor Function .