The first three terms all equal $1 / 3$, then the next five all equal $1 / 5$; more generally, for each $a=1,2, \ldots, 9$, the terms $1 /(2\lfloor\sqrt{a^{2}}\rfloor+1)$ to $1 /(2\lfloor\sqrt{a^{2}+2 a}\rfloor+1)$ all equal $1 /(2 a+1)$, and there are $2 a+1$ such terms. Thus our terms can be arranged into 9 groups, each with sum 1 , and only the last term $1 /(2\lfloor\sqrt{100}\rfloor+1)$ remains, so the answer is $9+1 / 21=190 / 21$.