The diagonal entries of a Hermitian tend to infinity, all entries of its inverse are zero Mathematically, suppose $A+\sigma I$ where $A \in \mathbb{S}^{n \times n}_{++}$ is positive, $I$ is the identity matrix, $\sigma$ is a scalar and $\sigma \rightarrow \infty$. Does it mean that $(A+\sigma I)^{-1}$ is a matrix where every entry is zero?  Intuitively, I know it's true. 
The problem is: how do we prove it? I was trying to prove by that the inverse is its adj over determinant.
 A: It is not true,  for finite $\sigma$, that every entry of $(\sigma I + A)^{-1}$ is zero; indeed, such a hypothesis doesn't really make sense, definition-wise, since it implies $0$ is the inverse of $\sigma I + A$, which is of course impossible.  
What is possible, and true for bounded operators $A$, is that $(\sigma I + A)^{-1} \to 0$ as $\sigma \to \infty$, and hence so do its entries if $A$ is identified with a matrix.  
A proof of this assertion follows.  Note that we require $A$ to have no special properties, other than the existence of $\Vert A \Vert$, which is guaranteed if $A$ is of finite size; $A$ need not be symmetric or positive.
We observe that
$(\sigma I + A)^{-1} = (\sigma(I + \sigma^{-1}A))^{-1} = \sigma^{-1}(I + \sigma^{-1}A)^{-1}; \tag{1}$
also, for $\sigma > \Vert A \Vert$ we have
$\Vert \sigma^{-1} A \Vert = \sigma^{-1} \Vert A \Vert < 1; \tag 2$
it then follows that
$(I + \sigma^{-1}A)^{-1} = (I - (-\sigma^{-1} A))^{-1}$
$= \displaystyle \sum_0^\infty (-1)^n \sigma^{-n}A^n = I - \sigma^{-1}A + \sigma^{-2}A^2 - \sigma^{-3}A^3 + \sigma^{-4}A^4 + \ldots, \tag 3$
where the series on the right converges absolutely and uniformly for $\sigma > \Vert A \Vert + \epsilon$, $\epsilon > 0$, since
$ \Vert \displaystyle \sum_0^\infty (-1)^n \sigma^{-n}A^n \Vert = \Vert I - \sigma^{-1}A + \sigma^{-2}A^2 - \sigma^{-3}A^3 + \sigma^{-4}A^4 + \ldots \Vert \le \sum_0^\infty \sigma^{-n} \Vert A \Vert^n, \tag 4$
and the latter series converges by virtue of (2); it is a geometric series with ratio $\sigma^{-1} \Vert A \Vert$; therefore we have
$\displaystyle \sum_0^\infty \sigma^{-n} \Vert A \Vert^n = \dfrac{1}{1 - \sigma^{-1} \Vert A \Vert}; \tag 5$
combining (3), (4) and (5) we write
$\Vert (I + \sigma^{-1}A)^{-1} \Vert \le  \dfrac{1}{1 - \sigma^{-1} \Vert A \Vert}; \tag 6$
we note that $1 / (1 - \sigma^{-1} \Vert A \Vert)$ is monotonically decreasing as a function of $\sigma$; thus we may choose $\sigma_0 > 0$ and for $\sigma > \sigma_0$,
$\Vert (I + \sigma^{-1}A)^{-1} \Vert \le  \dfrac{1}{1 - \sigma^{-1} \Vert A \Vert} <  \dfrac{1}{1 - \sigma_0^{-1} \Vert A \Vert}; \tag 7$
then from (1),
$\Vert (\sigma I + A)^{-1} \Vert =  \sigma^{-1} \Vert (I + \sigma^{-1}A)^{-1}) \Vert  < \sigma^{-1}  \dfrac{1}{1 - \sigma_0^{-1} \Vert A \Vert} \to 0 \; \text{as} \; \sigma \to \infty; \tag{8}$
since $\Vert (\sigma I + A)^{-1} \Vert \to 0$ as $\sigma \to \infty$, we also see that
$(\sigma I + A)^{-1} \to 0 \; \text{as} \; \sigma \to \infty, \tag 9$
which in turn implies the individual entries of $(\sigma I + A)^{-1}$ approach $0$ as well.
Note:  Careful inspection of the above argument indicates that even if $A$ is a continuous operator on an infinite-dimensional normed space, the result $(\sigma I + A)^{-1} \to 0$ still binds.  End of Note.
A: No, because after you take limit, there is no $\sigma$. But it is true that, as $\sigma\to\infty$, the entries of $(A+\sigma I)^{-1}$ go to zero. 
You can make it a proof by noting that each entry of the adjugate matrix of $A+\sigma I$ is of order $\sigma^{n-1}$, which $\det(A+\sigma I)$ is of order $\sigma^n$. So the entries of $(A+\sigma I)^{-1}$ are of order $1/\sigma$. 
Another way, if you write $B_\sigma=A+\sigma I$, and  $C_\sigma=B_\sigma^{-1}$, is to note that $C_\sigma^*C_\sigma=(B_\sigma^*B_\sigma)^{-1}$, and that the eigenvalues of the inverse are the inverses of the eigenvalues. Then, if $\lambda_1(X),\ldots,\lambda_n(X)$ denote the eigenvalues of $X$, counting multiplicites, 
\begin{align}
\sum_{k,j}|(C_{\sigma})_{kj}|^2&=\operatorname{Tr}(C_\sigma^*C_\sigma)
=\sum_{j=1}^n\lambda_j(C^*_\sigma C_\sigma)=\sum_j\frac1{\lambda_j(B_\sigma^* B_\sigma)}\\ \ \\
&\leq n\max\left\{\frac1{\lambda_j(B_\sigma^*B_\sigma)}:\ j=1,\ldots,n\right\}\\ \ \\
&= \frac n{\min\left\{\lambda_j(B_\sigma^*B_\sigma):\ j=1,\ldots,n\right\}}\\ \ \\
\end{align}
Now, if $\lambda$ is an eigenvalue of $B_\sigma^*B_\sigma$, there exists $v$ with $\|v\|=1$ and $B_\sigma^*B_\sigma v=\lambda v$. Suppose that $v=(v_1,\cdots)$ with $v_1\geq 1/\sqrt n$ (if it is not the first entry, it will be another one; I'm using the first one for simpler notation). We may assume $a>0$ because otherwise we may use $-v$. Then 
$$
\lambda=\langle\lambda v,v\rangle=\langle B_\sigma^*B_\sigma v,v\rangle=\langle B_\sigma v,B_\sigma v\rangle=\|B_\sigma v\|^2\geq\sigma^2 v_1^2\geq\sigma^2/n^2.
$$
Then 
$$
\min\left\{\lambda_j(B_\sigma^*B_\sigma):\ j=1,\ldots,n\right\}\geq\frac{\sigma^2}{n^2}.
$$
Combining with the previous estimate, we get 
$$
|(C_\sigma)_{st}|^2\leq \sum_{k,j}|(C_{\sigma})_{kj}|^2\leq \frac{n^3}{\sigma^2}
$$
