Calculation of the differential of the entropy In this review (for those who wants a precise reference see page 8 eq 21), the Author says that:
\begin{equation*}
S=-\sum_{i}P\left(i\right)\ln P\left(i\right)
\end{equation*}
and using the Maxwell-Boltzmann distribution
\begin{equation*}
P\left(i\right)=P_{MB}\left(i\right)+dP\left(i\right)
\end{equation*}
\begin{equation*}
dS=-\sum_{i}dP\left(i\right)\ln P_{MB}\left(i\right)=-\sum_{i}dP\left(i\right)\ln e^{-\beta E\left(i\right)}=\beta\sum_{i}dP\left(i\right)E\left(i\right)
\end{equation*}
However I tried to do the detailed calculation but here is what I get:
\begin{equation*}
dS=-\sum_{i}dP\left(i\right)\ln\left[P\left(i\right)\right]-\sum_{i}P\left(i\right)\frac{1}{P\left(i\right)}dP\left(i\right)=-\sum_{i}dP\left(i\right)\ln P\left(i\right)-\sum_{i}dP\left(i\right)=
\end{equation*}
\begin{equation*}
=-\sum_{i}dP\left(i\right)\ln\left[P_{MB}\left(i\right)+dP\left(i\right)\right]-\sum_{i}P\left(i\right)+\sum_{i}P_{MB}\left(i\right)
\end{equation*}
Can someone help me to get the right result? 
Thanks!
 A: Basically you just missed the fact that $\sum_{i} dP(i) =0$. (You also forgot that when you taylor expand to get $f(x_0+\delta x)$ you evaluate the derivatives in the expansion at $x_0$, not $x_0+\delta x$.),  I will give a full derivation anyway.
As you say, we have 
\begin{equation*}
S=-\sum_{i}P\left(i\right)\ln P\left(i\right)
\end{equation*}
Now we assume a change in the probability from $P_{MB}$:
\begin{equation*}
P\left(i\right)=P_{MB}\left(i\right)+dP\left(i\right)
\end{equation*}
The the entropy becomes
\begin{equation*}
\tilde{S}=-\sum_{i}(P_{MB}\left(i\right)+dP\left(i\right))\ln (P_{MB}\left(i\right)+dP\left(i\right)).
\end{equation*}
Thus we have a sum of terms of the form $(A+dA)\log(A+dA)$. To expand this, let's first note that 
\begin{equation}
\log(A+dA) = \log(A(1+\frac{dA}{A}))
=\log(A)+\log(1+dA/A)=\log(A)+ dA/A,
\end{equation}
where to get the last expression, we used $\log(1+x) \approx x$ for $x\ll 1$.
Then
\begin{equation}
(A+dA)\log(A+dA) = (A+dA)(\log(A)+ dA/A)=A\log A + dA \log A +dA,
\end{equation}
where we have dropped terms of order $(dA)^2$.
Our expression for the entropy $\tilde{S}$ becomes
\begin{equation*}
\tilde{S}=-\sum_{i}P_{MB}(i)\log(P_{MB}(i)) +dP(i) \log(P_{MB}(i)) +dP(i).
\end{equation*}
We recognize that the first term in this sum gives the original entropy. Thus the change in entropy $dS$ is given by 
\begin{equation}
\begin{aligned}
dS &= -\sum_{i}dP(i) \log(P_{MB}(i)) +dP(i) \\
&= -\sum_{i}dP(i) \log(P_{MB}(i))-\sum_{i}dP(i).
\end{aligned}
\end{equation}
Since total probability must be $1$, we have that $\sum_{i}dP(i)=0$. Also, by definition of the Maxwell-Boltzmann distribution, $ \log(P_{MB}(i)) = - \beta E(i)$. Plugging these two identities into our expression for $dS$ yields
\begin{equation}
dS = \beta \sum_{i}  E(i)dP(i)
\end{equation}
