Assume $k \ll 1$ then $W_{t+1}-W_t\approx \frac{dW_t}{dt}=kAe^{-kt}$. This is an over estimate of the difference, with absolute value of the error bounded by $\varepsilon(t)=k^2Ae^{-kt}/2$

If you want it exactly though:

$$ W_{t+1}-W_t=Ae^{-kt}-Ae^{-k(t+1)}=Ae^{-kt}(1-e^{-k}) $$

Assume $k \ll 1$ then $W_{t+1}-W_t\approx \frac{dW_t}{dt}=kAe^{-kt}$. This is an over estimate of the difference, with absolute value of the error bounded by $\varepsilon(t)=k^2Ae^{-kt}/2$

If you want it exactly though:

$$ W_{t+1}-W_t=Ae^{-kt}-Ae^{-k(t+1)}=Ae^{-kt}(1-e^{-k}) $$

An Octave session illustrating this is shown:

>> k=0.01;A=1;
>>
>> # Define W
>> W = @(t,A,k) A.*(1-exp(-k.*t));
>>
>> #calculate W_6-W_5
>> W5=W(5,A,k)
W5 =  0.048771
>> W6=W(6,A,k)
W6 =  0.058235
>> W6-W5
ans =  0.0094649
>>
>> #now calc approx difference
>> k*exp(-k*5)
ans =  0.0095123
>>
>> #now calc exact difference
>> A.*exp(-k.*t).*(1-exp(-k))
ans =  0.0094649
>>

Assume $k \ll 1$ then $W_{t+1}-W_t\approx \frac{dW_t}{dt}=kAe^{-kt}$.

If you want it exactly though:

$$ W_{t+1}-W_t=Ae^{-kt}-Ae^{-k(t+1)}=Ae^{-kt}(1-e^{-k}) $$

Assume $k \ll 1$ then $W_{t+1}-W_t\approx \frac{dW_t}{dt}=kAe^{-kt}$. This is an over estimate of the difference, with absolute value of the error bounded by $\varepsilon(t)=k^2Ae^{-kt}/2$

If you want it exactly though:

$$ W_{t+1}-W_t=Ae^{-kt}-Ae^{-k(t+1)}=Ae^{-kt}(1-e^{-k}) $$