<s>[INST] Looking for "an easy to understand" proof for following Power series I'm looking for proof for the following Power series
$exp(X) = \sum_{k=0}^{n} \frac{X^{k}}{k!}$
If X is $A_{nxn}$ matrix, then prove the series is converge 
Given
$\exp(X) = \sum_{k=0}^{n} \frac{X^{k}}{k!}$
$\sin(X) = \sum_{k=0}^{n} (-1)^{k}\frac{X^{2k}}{2k!}$
$\cos(X) = \sum_{k=0}^{n} (-1)^{k}\frac{X^{2k+1}}{(2k+1)!}$
$\cosh(X) = \sum_{k=0}^{n} \frac{X^{2k}}{(2k)!}$
$\sinh(X) = \sum_{k=0}^{n} \frac{X^{2k+1}}{(2k+1)!}$
Here is what I try so far, and I'm looking for "easy to understand" proof
Let $ X = 
    \begin{bmatrix}
    0 & -\beta \\
    \beta & 0 
    \end{bmatrix}
$
$
    \exp(X) = 
        \begin{bmatrix}
        1 & 0 \\
        0 & 1 
        \end{bmatrix}
        +
        \begin{bmatrix}
        0 & -\beta \\
        \beta & 0 
        \end{bmatrix}
        +
        \frac{1}{2!}
        \begin{bmatrix}
        -\beta^{2} & 0 \\
        0 & \beta^{2} 
        \end{bmatrix}
        +
        \frac{1}{3!}
        \begin{bmatrix}
        0 & \beta^{3} \\
        -\beta^{3} & 0 
        \end{bmatrix}
        +
        \frac{1}{4!}
        \begin{bmatrix}
        \beta^{4} & 0 \\
         0 & \beta^{4} 
        \end{bmatrix}
        +
        \frac{1}{5!}
        \begin{bmatrix}
        0 & -\beta^{5}\\
        \beta^{5} & 0 
        \end{bmatrix}
$
$
    \exp(X) = 
        \begin{bmatrix}
        1 + \frac{1}{2!}(-\beta^{2}) + \frac{1}{4!}\beta^{4} + ... & -\beta + \frac{1}{3!}\beta^{3} + ...\\
        \beta + \frac{1}{3!}-\beta^{3} + ... & 1 + \frac{1}{2!}\beta^{2} + \frac{1}{4!}\beta^{4} + ...
        \end{bmatrix}
        = 
        \begin{bmatrix}
        \sin(\beta) & -\cos(\beta)\\
        \cos(\beta) & \sin(\beta)
        \end{bmatrix}
$
 [/INST] Better to proceed as follows.
Consider for example the series $\sum_{k=0}^\infty\frac1{k!}t^kA^k$, say with $t$ complex or if you prefer take $t\in\mathbb R$. You can verify that the all entries of the series are power series.
So, all that matters for convergence is the radius of convergence. It is easy to show that the radius of convergence of the series of numbers $\sum_{k=0}^\infty\frac1{k!}t^k\|A^k\|$ is infinity, and so the same goes for each entry of the original series. 
Conclusion: each entry of $\sum_{k=0}^\infty\frac1{k!}t^kA^k$ converges, for all values of $t$, and you can take say $t=1$.
</s>