Specific case of Sherman–Morrison formula I want to find an easy solution to a matrix inverse for a very special case.
My matrix is of the form: $$D+vv^{T}$$
I know that I can use Sherman–Morrison formula to get
$$(D+vv^{T})^{-1}=D^{-1}-{D^{-1} vv^{T} D^{-1} \over 1+v^{T}D^{-1}v}$$
My D is actually of a very simple diagonal form. It is a diagonal matrix with all elements $1-s^2$. Thus, it looks like this
\begin{pmatrix} 1-s^2 & 0 & 0 \\
                  0 &  1-s^2 & 0 \\
                  0 & 0 &  1-s^2 \end{pmatrix}
and can be represented as $$D=(1-s^2)I$$ so that I can derive $$D^{-1}=\frac{1}{1-s^2}I$$ with $s$ being a scalar (between 0 and 1). 
I actually have an additional information that all elements of the vector $v$ would be $s$. So actually $v=s*u$ with $u=(1,\dots,n)^T$.
From here, I kind of struggle to simplify the Sherman–Morrison formula. 
1) Is the following correct?
$$\frac{1}{1-s^2}I-{\frac{1}{1-s^2}I vv^{T} \frac{1}{1-s^2}I \over 1+v^{T} \frac{1}{1-s^2}I v}$$
$$\frac{1}{1-s^2}I-{\frac{1}{1-s^2}I s^2uu^{T} \frac{1}{1-s^2}I \over 1+su^{T} \frac{1}{1-s^2}I su}$$
$$\frac{1}{1-s^2}I-\frac{s^2}{1-2s^2+s^4} {I uu^{T} I \over 1+\frac{s^2}{1-s^2} u^{T}Iu}$$
$$\frac{1}{1-s^2}I-\frac{s^2}{1-2s^2+s^4} {uu^{T} \over 1+\frac{s^2}{1-s^2} n}$$
2) Do you see a chance to simplify it even further?

\begin{align} 
\frac1{1-s^2}\cdot I-\frac{s^2}{({1-s^2)(1+s^2(n-1))}}\cdot\mathbf{1}
,
\end{align}  
that is, all diagonal elements are
\begin{align}
\frac{1+s^2(n-2)}{(1-s^2)(1+s^2(n-1))} 
,
\end{align}  
and all the other elements are
\begin{align}
-\frac{s^2}{(1-s^2)(1+s^2(n-1))}
.
\end{align}
