There are two fixes: either use a constant voltage (CV) finish, or use a much lower current to charge. The reason you are seeing this is iR drop. There is some finite ESR to the cell which the current flows across. If that ESR is 50 mOhm and you are charging with 200 mA you should expect a 10 mV drop in the potential after the charge current goes away.

Using a much lower charging current reduces the iR drop, but means much larger charging times which is why CV phases are used: so you can charge relatively quickly, but still get all the capacity into the cell. If you don't want to use a CV phase then the only other option to avoid the iR drop issue is lower charging current.

Thank you Jacob for your answer !

I read in some scientific paper that we can decrease this voltage drop by using the lowest impedance which we can get at high frequency. But do you think even if I found the right frequency of pulsing to have the minimum impedance, will that decrease a lot this drop?

I want to say no, because the ESR is what it is and so that relaxation is always the same regardless of current pulse duration, but there are also capacitive effects to deal with and a slower pulse that charges the capacitance but doesn't charge the cell may show a bigger (and slower) voltage relaxation after the current pulse is removed. I would think, however, that a larger fraction of the charge that passes in your pulse goes to charging the battery if the pulse is longer, and so even with a larger (potential) voltage drop you still more effectively charge the cell with longer pulses.

Whether that would be useful for you really depends on what you are trying to do.