Why is fluoride the most nucleophilic halide in aprotic solvents? I am confused on the topic of electronegative nucleophiles, and polar aprotic solvents. 

From my understanding, the more electronegative a nucleophile is, the weaker of a nucleophile it will be, because atoms will be less likely to share its electron density. 

However, in a polar aprotic solvent, because H-bonding cannot occur, it seems as if the trend of nucleophilicity is such that $\ce{F- &gt; Cl- &gt; Br- &gt; I-}$.

Can someone please clarify how this is possible? I would have expected $\ce{F-}$ to be the LEAST nucleophilic nucleophile in a polar aprotic solvent, because it is the most electronegative atom. Therefore according to the first point I learned, I thought it should be the weakest nucleophile. 

All this is what I've learned through reading Kaplan's Organic Chemistry Review.
When talking about elements of a group, the size of the atom is more effective than the electronegativity factor.
As $\ce{F-}$ size is too short so its electron density will be much higher than other hallogens.

In aprotic solvent the order of nucleophillicity:
$\ce{F- &gt; Cl- &gt; Br- &gt;I-}$(No hydration occurs) In protic solvant(like $\ce{H2O}$) its reversed because of hydration of ions having much electron density.