Let $I$ be defined as
$$I=\displaystyle\int_0^1\Big\lfloor\frac1x\Big\rfloor\Big\{\frac1x\Big\}dx$$
I found that
$$I=\displaystyle\sum_{k=1}^\infty\left(k\ln\left(\frac{k+1}{k}\right)-\frac{k}{k+1}\right)$$
which I'm not sure if converges or not. Any thoughts??
Integral of Integral part times Fractional part
integrationsequences-and-series
Best Answer
For $0 < x \le 1$ is $$ \ln (1+x) = x - \frac 12 x^2 + \frac 13 x^3 - \frac 14 x^4 + \cdots $$ an alternating series with terms of decreasing magnitude, and therefore $$ \ln (1+x) > x - \frac 12 x^2 \, . $$ (Actually that inequality holds for all $x > 0$, but we don't need that here.)
It follows that $$ k\ln\left(1+\frac{1}{k}\right)-\frac{k}{k+1} \ge k \left(\frac 1k - \frac{1}{2k^2}\right) -\frac{k}{k+1} = \frac{k-1}{2k(k+1)} \sim \frac{1}{2k} $$ for $k \to \infty$, so that the series and the integral are divergent.