We know that, if a matrix $A \in \mathbb{M}_N(\mathbb{R})$ is invertible, its inverse can be written as:
$${A^{ - 1}} = \frac{1}{{\det \left( A \right)}}{C^T},$$
where $$C = \left[ {\begin{array}{*{20}{c}}{{C_{11}}}&{{C_{12}}}& \cdots &{{C_{1N}}}\\{{C_{21}}}&{{C_{22}}}& \cdots &{{C_{2N}}}\\ \vdots & \vdots & \ddots & \vdots \\{{C_{N1}}}&{{C_{N2}}}& \cdots &{{C_{NN}}}\end{array}} \right]$$
is the cofactor matrix, with the $(i,j)$-cofactor defined as: $${C_{ij}} = {\left( { - 1} \right)^{i + j}}{M_{i,j}},$$
being $M_{i,j}$ the $(i,j)$-minor, which is by definition the determinant of the submatrix formed by deleting the i-th row and j-th column of $A$.
Now, let's consider:
$${A_N} = {\left[ {\begin{array}{*{20}{c}}1&{ - 1/2}&0& \cdots &0&0\\{ - 1/2}&1&{ - 1/2}& \cdots &0&0\\0&{ - 1/2}&1& \ddots &0&0\\ \vdots & \vdots & \ddots & \ddots & \vdots & \vdots \\0&0&0& \cdots &1&{ - 1/2}\\0&0&0& \cdots &{ - 1/2}&1\end{array}} \right]_{N \times N}}.$$
As you are only interested in the first entry of $A^{-1}_{N}$, we are done if we can find $C_{11}$ and $\det \left( A_{N} \right)$, as $${\left( {A_N^{ - 1}} \right)_{1,1}} = \frac{{{C_{11}}}}{{\det \left( {{A_N}} \right)}}.$$
Taking $a_{N}=\det \left( {{A_N}} \right)$ and applying Laplace's expansion along the first row, we get:
$$\begin{array}{c}{a_N} = \det {\left[ {\begin{array}{*{20}{c}}1&{ - 1/2}&0&0& \cdots &0&0\\{ - 1/2}&1&{ - 1/2}&0& \cdots &0&0\\0&{ - 1/2}&1&{ - 1/2}& \cdots &0&0\\ \vdots & \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\0&0&0&0& \cdots &1&{ - 1/2}\\0&0&0&0& \cdots &{ - 1/2}&1\end{array}} \right]_{\left( {N - 1} \right) \times \left( {N - 1} \right)}}\\ + \frac{1}{2}\det {\left[ {\begin{array}{*{20}{c}}{ - 1/2}&{ - 1/2}&0& \cdots &0&0\\0&1&{ - 1/2}& \cdots &0&0\\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\0&0&0& \cdots &1&{ - 1/2}\\0&0&0& \cdots &{ - 1/2}&1\end{array}} \right]_{\left( {N - 1} \right) \times \left( {N - 1} \right)}}.\end{array}$$
But:
$$\det {\left[ {\begin{array}{*{20}{c}}1&{ - 1/2}&0&0& \cdots &0&0\\{ - 1/2}&1&{ - 1/2}&0& \cdots &0&0\\0&{ - 1/2}&1&{ - 1/2}& \cdots &0&0\\ \vdots & \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\0&0&0&0& \cdots &1&{ - 1/2}\\0&0&0&0& \cdots &{ - 1/2}&1\end{array}} \right]_{\left( {N - 1} \right) \times \left( {N - 1} \right)}}=a_{N-1},$$
and, again taking Laplace's expansion along the first row:
$$\begin{array}{l}\det {\left[ {\begin{array}{*{20}{c}}{ - 1/2}&{ - 1/2}&0& \cdots &0&0\\0&1&{ - 1/2}& \cdots &0&0\\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\0&0&0& \cdots &1&{ - 1/2}\\0&0&0& \cdots &{ - 1/2}&1\end{array}} \right]_{\left( {N - 1} \right) \times \left( {N - 1} \right)}} = \\ = - \frac{1}{2}\det {\left[ {\begin{array}{*{20}{c}}1&{ - 1/2}& \cdots &0&0\\ \vdots & \vdots & \cdots & \vdots & \vdots \\0&0& \cdots &1&{ - 1/2}\\0&0& \cdots &{ - 1/2}&1\end{array}} \right]_{\left( {N - 2} \right) \times \left( {N - 2} \right)}}\\{\rm{ }} + \frac{1}{2}\det {\left[ {\begin{array}{*{20}{c}}0&{ - 1/2}& \cdots &0&0\\ \vdots & \vdots & \cdots & \vdots & \vdots \\0&0& \cdots &1&{ - 1/2}\\0&0& \cdots &{ - 1/2}&1\end{array}} \right]_{\left( {n - 2} \right) \times \left( {N - 2} \right)}}\\ = - \frac{1}{2}{a_{N - 2}} + \frac{1}{2} \times 0.\end{array}$$
Hence, we get the following linear difference equation for $a_{N}=\det \left( {{A_N}} \right)$:
$${a_N} = {a_{N - 1}} - \frac{1}{4}{a_{N - 2}}.$$
There are many different techniques to solve linear difference equations, but I'm going to use my favorite, which is based uniquely on linear algebra.
To get started, let's take $b_N=a_{N-1}$, which allows us to write the second order linear difference equation as a system of two first order linear equations:
$$\left\{ \begin{array}{l}{a_N} = {a_{N - 1}} - \frac{1}{4}{b_{N - 1}}\\{b_N} = {a_{N - 1}}\end{array} \right. \Leftrightarrow \underbrace {\left[ {\begin{array}{*{20}{c}}{{a_N}}\\{{b_N}}\end{array}} \right]}_{{X_N}} = \underbrace {\left[ {\begin{array}{*{20}{c}}1&{ - 1/4}\\1&0\end{array}} \right]}_M\underbrace {\left[ {\begin{array}{*{20}{c}}{{a_{N - 1}}}\\{{b_{N - 1}}}\end{array}} \right]}_{{X_{N - 1}}}.$$
At this point, let's breathe and think what we can get from the last expression. We actually know $X_2$:
$${X_2} = \left[ {\begin{array}{*{20}{c}}{{a_2}}\\{{b_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{a_2}}\\{{a_1}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\det \left( {{A_2}} \right)}\\{\det \left( {{A_1}} \right)}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\left| {\begin{array}{*{20}{c}}1&{ - 1/2}\\{ - 1/2}&1\end{array}} \right|}\\{\left| 1 \right|}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{3/4}\\1\end{array}} \right].$$
Therefore, we have $X_3=M X_2$, $X_4=M X_3=M^2 X_2$, ... and this inductively leads us to the following expression
$${X_N} = {M^{N - 2}}{X_2}.$$
If we are able to find $M^{N-2}$, then we can find the general expression for $X_N$ and, thus, arrive at the solution, $a_N$, to our original second order linear difference equation.
Let's take the Jordan decomposition of $M$, $M=S J S^{-1}$. Then:
$${M^{N - 2}} = \underbrace {SJ{S^{ - 1}}SJ{S^{ - 1}} \ldots SJ{S^{ - 1}}}_{(N-2){\rm{ times}}} = SJ\left( {{S^{ - 1}}S} \right)J\left( {{S^{ - 1}}S} \right) \ldots \left( {{S^{ - 1}}S} \right)J{S^{ - 1}} = S{J^{N - 2}}S.$$
To find, $S$ and $J$ we need to find the eigenvalues and eigenvectors of $M$. If $M$ is diagonalizable, then $J$ will be a diagonal matrix whose entries are the eigenvalues of $M$, and $S$ will have the corresponding eigenvectors as columns. Otherwise, if $M$ has only one eigenvalue, say $\lambda$, of algebraic multiplicity 2 and its eigenspace is of dimension 1 (that is, we can only "extract" one eigenvector associated to $\lambda$) then, as the matrix $M$ is of order 2, we know that $J$ will have the form:
$$J = \left[ {\begin{array}{*{20}{c}}\lambda &1\\0&\lambda \end{array}} \right],$$
and $S$ will have the only eigenvector we could find as its first column and a generalized eigenvector as its second column.
The characteristic polynomial associated to $M$ is:
$$p\left( \lambda \right) = {\left( {\lambda - \frac{1}{2}} \right)^2},$$
so $M$ has only one eigenvalue of algebraic multiplicity 2: $\lambda = 1/2$.
After a few calculations, we find that the eigenspace associated to $\lambda = \frac{1}{2}$ is:
$${E_\lambda } = N\left( {M - \frac{1}{2}I} \right) = \left\{ {\alpha \in \mathbb{R}:\alpha \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]} \right\},$$
so $\dim \left( {{E_\lambda }} \right) = 1$ and we only get one eigenvector; let's take: $${v_\lambda } = {\left[ {\begin{array}{*{20}{c}}1&2\end{array}} \right]^T}.$$ We now need to find a generalized eigenvector and that we can do by solving the following linear system:
$$\left( {M - \frac{1}{2}I} \right)x = {v_\lambda }.$$
The solution for this system is: $$x = {\left[ {\begin{array}{*{20}{c}}3&2\end{array}} \right]^T},$$
and so we get:
$$S = \left[ {\begin{array}{*{20}{c}}1&3\\2&2\end{array}} \right]; J = \left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&1\\0&{\frac{1}{2}}\end{array}} \right].$$
Doing some more calculation we can easily find:
$${S^{ - 1}} = \left[ {\begin{array}{*{20}{r}}{ - \frac{1}{2}}&{\frac{3}{4}}\\{\frac{1}{2}}&{ - \frac{1}{4}}\end{array}} \right].$$
Now, we only need to compute $J^{N-2}$. Fortunately for us it is a well known fact (you can prove it by induction) that:
$${\left[ {\begin{array}{*{20}{r}}\lambda &1\\0&\lambda \end{array}} \right]^N} = \left[ {\begin{array}{*{20}{c}}{{\lambda ^N}}&{N{\lambda ^{N - 1}}}\\0&{{\lambda ^N}}\end{array}} \right],$$
so we get:
$${J^{N - 2}} = \left[ {\begin{array}{*{20}{c}}{\frac{1}{{{2^{N - 2}}}}}&{\left( {N - 2} \right)\frac{1}{{{2^{N - 3}}}}}\\0&{\frac{1}{{{2^{N - 2}}}}}\end{array}} \right].$$
After some really messy calculations we arrive at:
$${X_N} = \left[ {\begin{array}{*{20}{c}}{{a_N}}\\{{b_N}}\end{array}} \right] = {M^{N - 2}}{X_2} = S{J^{N - 2}}{S^{ - 1}}{X_2} = \left[ {\begin{array}{*{20}{c}}{\frac{{N + 1}}{{{2^N}}}}\\{\frac{N}{{{2^{N - 1}}}}}\end{array}} \right],$$
and we can conclude that:
$${a_N} = \det \left( {{A_N}} \right) = \frac{{N + 1}}{{{2^N}}}.$$
Let's return to our initial problem of finding ${\left( {A_N^{ - 1}} \right)_{1,1}}$.
We are now only missing $C_{11}$.
$${M_{1,1}} = \det {\left[ {\begin{array}{*{20}{c}}1&{ - 1/2}&0&0& \cdots &0&0\\{ - 1/2}&1&{ - 1/2}&0& \cdots &0&0\\0&{ - 1/2}&1&{ - 1/2}& \cdots &0&0\\ \vdots & \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\0&0&0&0& \cdots &1&{ - 1/2}\\0&0&0&0& \cdots &{ - 1/2}&1\end{array}} \right]_{\left( {N - 1} \right) \times \left( {N - 1} \right)}} = {a_{N - 1}} = \frac{{N}}{{{2^{N-1}}}}.$$
Hence:
$${C_{11}} = {\left( { - 1} \right)^{1 + 1}}{M_{1,1}} = \frac{N}{{{2^{N - 1}}}},$$
and we can finally conclude that:
$${\left( {A_N^{ - 1}} \right)_{1,1}} = \frac{{{C_{11}}}}{{\det \left( {{A_N}} \right)}} = \frac{N}{{{2^{N - 1}}}}\frac{{{2^N}}}{{N + 1}} = \frac{{2N}}{{N + 1}}, Q.E.D.$$
