$$I=\int_0^1\left\{\frac 1x\right\}dx=\int_1^\infty\frac{\{u\}}{u^2}du=\sum_{k=1}^\infty\int_0^1\frac{\{v+k\}}{(v+k)^2}dv=\sum_{k=1}^\infty\int_0^1\frac{v}{(v+k)^2}dv=\sum_{k=1}^\infty\ln\left(\frac{k+1}k\right)+\frac k{k+1}-1$$ and I believe the integral should converge but neither parts of this series converge from what ive calculated: $$\sum_{k=1}^\infty\ln\left(\frac{k+1}k\right)=\ln\left(\prod_{k=1}^\infty\frac{k+1}k\right)=\lim_{n\to\infty}\ln\left(\frac{(n+1)!}{n!}\right)=\lim_{n\to\infty}\ln(n+1)\to\infty$$
my reasoning is that if the first substitution is valid: $$\int_1^\infty\frac{\{u\}}{u^2}du\le\int_1^\infty\frac{du}{u^2}=\left[\frac{1}{u}\right]_\infty^1=1$$
Thank you for all the comments and answers, using them I have written the sum as: $$1+lim_{n\to\infty}\left[\ln(n)-\text{H}_n\right]$$ which as has been pointed out is a known value of $1-\gamma$
To answer your first question, the initial integral must converge. Write it as $$I(x)=\int_{1-x}^1\left\{\frac1t\right\}dt.$$ Since $0\le\left\{\frac1x\right\}<1$ for all $x$, we know that $I(x)$ is monotonically increasing and bounded (on $[0,1]$, and hence must converge).
As achille hui stated, the positive part of your summation diverges -- but so does the negative part! This by no means implies divergence.
In fact, the Euler Mascheroni constant is defined to be equal to this difference between the logarithmic function and the harmonic series sum. The integral, when evaluated , should be $1-\gamma.$